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NCERT Solutions for Class 12 Maths Chapter 7 Integrals Miscellaneous Exercise

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NCERT Solutions for Maths Integrals Miscellaneous Exercise Class 12 Chapter 7 - Free PDF Download

NCERT Solutions for Class 12 Maths Chapter 7 Integrals includes solutions to all Miscellaneous Exercise problems. Miscellaneous Exercise Class 12 Chapter 7 is based on the concepts presented in Maths Chapter 7. To perform well on the board exam, download the NCERT Solutions for Class 12 Maths Integrals Miscellaneous Exercise in PDF format and practice them offline. Students can download the revised Class 12 Maths NCERT Solutions from our page, which is prepared so that you can understand it easily.

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Class 12 Chapter 7 Maths Miscellaneous Exercise Solutions are aligned with the updated CBSE guidelines for Class 12, ensuring students are well-prepared for exams. Access the Class 12 Maths Syllabus here.

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Access NCERT Solutions for Class 12 Maths Chapter 7 Integrals

Miscellaneous Exercise

Integrate the functions in Exercises 1 to 23.

1. $\dfrac{1}{x-{{x}^{3}}}$

Ans: Given $\dfrac{1}{x-{{x}^{3}}}$

So, $\dfrac{1}{x-{{x}^{3}}}=\dfrac{1}{x\left( 1-{{x}^{2}} \right)}$

$=\dfrac{1}{x\left( 1-x \right)\left( 1+x \right)}$

Let $\dfrac{1}{x\left( 1-x \right)\left( 1+x \right)}=\dfrac{A}{x}+\dfrac{B}{\left( 1-x \right)}+\dfrac{C}{1+x}$                 …….(1)

$\Rightarrow 1=A\left( 1-{{x}^{2}} \right)+Bx\left( 1+x \right)+Cx\left( 1-x \right)$

$\Rightarrow 1=A-A{{x}^{2}}+Bx+B{{x}^{2}}+Cx-C{{x}^{2}}$

On equating the coefficients of ${{x}^{2}},x$ and constant term –

$-A+B-C=0$              

$B+C=0$

$A=1$

Thus, we get $A=1,B=\dfrac{1}{2}$ and $C=-\dfrac{1}{2}$

By equation 1-

$\dfrac{1}{x\left( 1-x \right)\left( 1+x \right)}=\dfrac{1}{x}+\dfrac{1}{2\left( 1-x \right)}-\dfrac{1}{2\left( 1+x \right)}$

$\Rightarrow \int{\dfrac{1}{x\left( 1-x \right)\left( 1+x \right)}}dx=\int{\dfrac{1}{x}dx}+\dfrac{1}{2}\int{\dfrac{1}{\left( 1-x \right)}}dx-\dfrac{1}{2}\int{\dfrac{1}{\left( 1+x \right)}dx}$

$=\log x-\dfrac{1}{2}\log \left( 1-x \right)-\dfrac{1}{2}\log \left( 1+x \right)$

$=\log x-\dfrac{1}{2}\log {{\left( 1-x \right)}^{\dfrac{1}{2}}}-\dfrac{1}{2}\log {{\left( 1+x \right)}^{\dfrac{1}{2}}}$

$=\log \left( \dfrac{x}{{{\left( 1-x \right)}^{\dfrac{1}{2}}}{{\left( 1+x \right)}^{\dfrac{1}{2}}}} \right)+C$

$=\log {{\left( \dfrac{{{x}^{2}}}{\left( 1-{{x}^{2}} \right)} \right)}^{\dfrac{1}{2}}}+C$

$=\dfrac{1}{2}\log \left( \dfrac{{{x}^{2}}}{1-{{x}^{2}}} \right)+C$


2. $\dfrac{1}{\sqrt{x+a}+\sqrt{x+b}}$

Ans: Given $\dfrac{1}{\sqrt{x+a}+\sqrt{x+b}}$

$=\dfrac{1}{\sqrt{x+a}+\sqrt{x+b}}\times \dfrac{\sqrt{x+a}-\sqrt{x+b}}{\sqrt{x+a}-\sqrt{x+b}}$

$=\dfrac{\sqrt{x+a}-\sqrt{x+b}}{\left( x+a \right)-\left( x+b \right)}$

$=\dfrac{\sqrt{x+a}-\sqrt{x+b}}{a-b}$

$\Rightarrow \int{\dfrac{1}{\sqrt{x+a}+\sqrt{x+b}}}dx=\dfrac{1}{a-b}\int{\left( \sqrt{x+a}-\sqrt{x+b} \right)}dx$

$=\dfrac{1}{a-b}\left[ \dfrac{{{\left( x+a \right)}^{\dfrac{3}{2}}}}{\dfrac{3}{2}}-\dfrac{{{\left( x+b \right)}^{\dfrac{3}{2}}}}{\dfrac{3}{2}} \right]$

$=\dfrac{2}{3\left( a-b \right)}\left[ {{\left( x+a \right)}^{\dfrac{3}{2}}}-{{\left( x+b \right)}^{\dfrac{3}{2}}} \right]+C$


3. $\dfrac{1}{x\sqrt{ax-{{x}^{2}}}}$                    $\left[ \text{Hint: x=}\dfrac{a}{t} \right]$

Ans: Given

$\dfrac{1}{x\sqrt{ax-{{x}^{2}}}}$

Let $x=\dfrac{a}{t}$

On differentiating-

$dx=-\dfrac{a}{{{t}^{2}}}dt$

$\Rightarrow \int{\dfrac{1}{x\sqrt{ax-{{x}^{2}}}}dx=\int{\dfrac{1}{\dfrac{a}{t}\sqrt{a.\dfrac{a}{t}-{{\left( \dfrac{a}{t} \right)}^{2}}}}}\left( -\dfrac{a}{{{t}^{2}}}dt \right)}$

$=-\int{\dfrac{1}{at}.\dfrac{1}{\sqrt{\dfrac{1}{t}-\dfrac{1}{{{t}^{2}}}}}}dt$

$=-\dfrac{1}{a}\int{\dfrac{1}{\sqrt{\dfrac{{{t}^{2}}}{t}-\dfrac{{{t}^{2}}}{{{t}^{2}}}}}}dt$

$=-\dfrac{1}{a}\int{\dfrac{1}{\sqrt{t-1}}}dt$

$=-\dfrac{1}{a}\left[ 2\sqrt{t-1} \right]+C$

Substituting value of t-

$=-\dfrac{1}{a}\left[ 2\sqrt{\dfrac{a}{x}-1} \right]+C$

$=-\dfrac{2}{a}\left[ \dfrac{\sqrt{a-x}}{\sqrt{x}} \right]+C$

$=-\dfrac{2}{a}\left[ \sqrt{\dfrac{a-x}{x}} \right]+C$


4. $\dfrac{\text{1}}{{{\text{x}}^{\text{2}}}{{\left( {{\text{x}}^{\text{4}}}\text{+1} \right)}^{\dfrac{\text{3}}{\text{4}}}}}$. 

Ans:The given expression is, $\dfrac{1}{{{x}^{2}}{{\left( {{x}^{4}}+1 \right)}^{\dfrac{3}{4}}}}$. 

On multiplying and dividing by ${{x}^{-3}}$, the following can be obtained as shown below,

$\dfrac{{{x}^{-3}}}{{{x}^{2}}.{{x}^{-3}}{{\left( {{x}^{4}}+1 \right)}^{\dfrac{3}{4}}}}=\dfrac{{{x}^{-3}}{{\left( {{x}^{4}}+1 \right)}^{\dfrac{-3}{4}}}}{{{x}^{2}}-{{x}^{-3}}}=\dfrac{{{\left( {{x}^{4}}+1 \right)}^{\dfrac{-3}{4}}}}{{{x}^{5}}{{\left( {{x}^{4}} \right)}^{\dfrac{-3}{4}}}}=\dfrac{1}{{{x}^{5}}}{{\left( \dfrac{{{x}^{4}}+1}{{{x}^{4}}} \right)}^{\dfrac{-3}{4}}}=\dfrac{1}{{{x}^{5}}}{{\left( 1+\dfrac{1}{{{x}^{4}}} \right)}^{\dfrac{-3}{4}}}$ 

Now, consider $\dfrac{1}{{{x}^{4}}}=t$ 

$\therefore \dfrac{-4}{{{x}^{5}}}=\dfrac{dt}{dx}$

$\Rightarrow \dfrac{dx}{{{x}^{5}}}=\dfrac{-dt}{4}$

$\therefore \int{\dfrac{1}{{{x}^{2}}{{\left( {{x}^{4}}+1 \right)}^{\dfrac{3}{4}}}}dx}=\int{\dfrac{1}{{{x}^{5}}}{{\left( 1+\dfrac{1}{{{x}^{4}}} \right)}^{\dfrac{-3}{4}}}dx}=\dfrac{-1}{4}\int{{{\left( 1+t \right)}^{\dfrac{-3}{4}}}dt}$ 

$\Rightarrow \int{\dfrac{1}{{{x}^{2}}{{\left( {{x}^{4}}+1 \right)}^{\dfrac{3}{4}}}}dx}=\dfrac{-1}{4}\left[ \dfrac{{{\left( 1+t \right)}^{\dfrac{1}{4}}}}{\dfrac{1}{4}} \right]+C=-{{\left( 1+\dfrac{1}{{{x}^{4}}} \right)}^{\dfrac{1}{4}}}+C$, where $C$ is any arbitrary constant.


5. $\dfrac{\text{1}}{{{\text{x}}^{\dfrac{\text{1}}{\text{2}}}}\text{+}{{\text{x}}^{\dfrac{\text{1}}{\text{3}}}}}$. $\text{[}$Hint: $\dfrac{\text{1}}{{{\text{x}}^{\dfrac{\text{1}}{\text{2}}}}\text{+}{{\text{x}}^{\dfrac{\text{1}}{\text{3}}}}}\text{=}\dfrac{\text{1}}{{{\text{x}}^{\dfrac{\text{1}}{\text{3}}}}\left( \text{1+}{{\text{x}}^{\dfrac{\text{1}}{\text{6}}}} \right)}$ Put, $\text{x=}{{\text{t}}^{\text{6}}}$ $\text{]}$

Ans: The given expression is, $\dfrac{1}{{{x}^{\dfrac{1}{2}}}+{{x}^{\dfrac{1}{3}}}}$. 

Observe the given hint and obtain as shown below,

$\therefore \dfrac{1}{{{x}^{\dfrac{1}{2}}}+{{x}^{\dfrac{1}{3}}}}=\dfrac{1}{{{x}^{\dfrac{1}{3}}}\left( 1+{{x}^{\dfrac{1}{6}}} \right)}$ 

Consider $x={{t}^{6}}$ 

$\therefore x={{t}^{6}}\Rightarrow dx=6{{t}^{5}}$

$\therefore \int{\dfrac{1}{{{x}^{\dfrac{1}{2}}}+{{x}^{\dfrac{1}{3}}}}dx}=\int{\dfrac{1}{{{x}^{\dfrac{1}{3}}}\left( 1+{{x}^{\dfrac{1}{6}}} \right)}dx}=\int{\dfrac{6{{t}^{5}}}{{{t}^{2}}\left( 1+t \right)}dt}=6\int{\dfrac{{{t}^{3}}}{\left( 1+t \right)}dt}$ 

Now on dividing, we can obtain as shown below,

$\int{\dfrac{1}{{{x}^{\dfrac{1}{2}}}+{{x}^{\dfrac{1}{3}}}}dx}=6\int{\left\{ \left( {{t}^{2}}-t+1 \right)-\dfrac{1}{1+t} \right\}dt}$

$=6\int{\left[ \left( \dfrac{{{t}^{3}}}{3} \right)-\left( \dfrac{{{t}^{2}}}{2} \right)+t-\log \left| 1+t \right| \right]dt}$

$=2{{x}^{\dfrac{1}{2}}}-3{{x}^{\dfrac{1}{3}}}+6{{x}^{\dfrac{1}{6}}}-6\log \left( 1+{{x}^{\dfrac{1}{6}}} \right)+C$

$=2\sqrt{x}-3{{x}^{\dfrac{1}{3}}}+6{{x}^{\dfrac{1}{6}}}-6\log \left( 1+{{x}^{\dfrac{1}{6}}} \right)+C$, where $C$ is any arbitrary constant.


6. $\dfrac{\text{5x}}{\text{(x+1)}\left( {{\text{x}}^{\text{2}}}\text{+9} \right)}$.

Ans: The given expression is, $\dfrac{5x}{(x+1)\left( {{x}^{2}}+9 \right)}$. 

Now consider it as shown below,

$\therefore \dfrac{5x}{(x+1)\left( {{x}^{2}}+9 \right)}=\dfrac{A}{\left( x+1 \right)}+\dfrac{Bx+C}{\left( {{x}^{2}}+9 \right)}\,\,\,\,\,\,\,\,\,\,\,...\left( 1 \right)$

$\Rightarrow 5x=A\left( {{x}^{2}}+9 \right)+Bx+C\left( x+1 \right)$

$\Rightarrow 5x=A{{x}^{2}}+9A+B{{x}^{2}}+Bx+Cx+C$

On equating the coefficients of ${{x}^{2}},\,x$ and constant term, it can be obtained that,  

$A+B=0\,\,\,...\left( 2 \right)$ 

$B+C=5\,\,\,...\left( 3 \right)$

$9A+C=0\,\,\,...\left( 4 \right)$

And on solving these equations, the values of $A,\,B,\,C$ can be obtained as,

$A=\dfrac{-1}{2},\,B=\dfrac{1}{2},\,C=\dfrac{9}{2}$ respectively.

Now, from equation $\left( 1 \right)$ it can be clearly obtained that,

$\int{\dfrac{5x}{(x+1)\left( {{x}^{2}}+9 \right)}dx}=\int{\left\{ \dfrac{-1}{2\left( x+1 \right)}+\dfrac{x+9}{2\left( {{x}^{2}}+9 \right)} \right\}dx}$ 

$=\dfrac{-1}{2}\log \left| x+1 \right|+\dfrac{1}{2}\int{\dfrac{x}{{{x}^{2}}+9}dx}+\dfrac{9}{2}\int{\dfrac{1}{{{x}^{2}}+9}dx}$

$=\dfrac{-1}{2}\log \left| x+1 \right|+\dfrac{1}{4}\log \left| {{x}^{2}}+9 \right|+\left( \dfrac{9}{2} \right)\left( \dfrac{1}{3} \right){{\tan }^{-1}}\left( \dfrac{x}{3} \right)$

$=\dfrac{-1}{2}\log \left| x+1 \right|+\dfrac{1}{4}\log \left| {{x}^{2}}+9 \right|+\dfrac{3}{2}{{\tan }^{-1}}\left( \dfrac{x}{3} \right)+C$, where $C$ is any arbitrary constant.


7. $\dfrac{\text{sinx}}{\text{sin}\left( \text{x- }\!\!\alpha\!\!\text{ } \right)}$.

Ans: The given expression is, $\dfrac{\sin x}{\sin \left( x-\alpha  \right)}$. 

 Now, substitute $x-\alpha =t$ 

$\therefore dx=dt$

$\therefore \int{\dfrac{\sin x}{\sin \left( x-\alpha  \right)}dx}=\int{\dfrac{\sin \left( t+\alpha  \right)}{\sin t}dt}=\int{\dfrac{\sin t\cos \alpha +\cos tsin\alpha }{\sin t}dt}=\int{\cos \alpha +\cot tsin\alpha dt}$$\Rightarrow \int{\dfrac{\sin x}{\sin \left( x-\alpha  \right)}dx}=t\cos \alpha +\sin \alpha \log \left| \sin t \right|+{{C}_{1}}=\left( x-\alpha  \right)\cos \alpha +\sin \alpha \log \left| \sin \left( x-\alpha  \right) \right|+{{C}_{1}}$ 

$\Rightarrow \int{\dfrac{\sin x}{\sin \left( x-\alpha  \right)}dx}=x\cos \alpha +\sin \alpha \log \left| \sin \left( x-\alpha  \right) \right|-\alpha \cos \alpha +{{C}_{1}}=x\cos \alpha +\sin \alpha \log \left| \sin \left( x-\alpha  \right) \right|+C$

 where ${{C}_{1}},\,C$ are any arbitrary constants and $C={{C}_{1}}-\alpha \cos \alpha $.


8. $\dfrac{{{\text{e}}^{\text{5logx}}}\text{-}{{\text{e}}^{\text{4logx}}}}{{{\text{e}}^{\text{3logx}}}\text{-}{{\text{e}}^{\text{2logx}}}}$.

Ans: The given expression is, $\dfrac{{{e}^{5\log x}}-{{e}^{4\log x}}}{{{e}^{3\log x}}-{{e}^{2\log x}}}$. 

 $\therefore \dfrac{{{e}^{5\log x}}-{{e}^{4\log x}}}{{{e}^{3\log x}}-{{e}^{2\log x}}}=\dfrac{{{e}^{4\log x}}\left( {{e}^{\log x}}-1 \right)}{{{e}^{2\log x}}\left( {{e}^{\log x}}-1 \right)}={{e}^{2\log x}}={{x}^{2}}$

Now, integrate the given expression as shown below

$\therefore \int{\dfrac{{{e}^{5\log x}}-{{e}^{4\log x}}}{{{e}^{3\log x}}-{{e}^{2\log x}}}dx}=\int{{{x}^{2}}dx}=\dfrac{{{x}^{3}}}{3}+C$, where $C$ is any arbitrary constant. 


9. $\dfrac{\text{cosx}}{\sqrt{\text{4-si}{{\text{n}}^{\text{2}}}\text{x}}}$.

Ans: The given expression is, $\dfrac{\cos x}{\sqrt{4-{{\sin }^{2}}x}}$. 

 Now, substitute $\sin x=t$ 

$\therefore \cos xdx=dt$

$\therefore \int{\dfrac{\cos x}{\sqrt{4-{{\sin }^{2}}x}}dx}=\int{\dfrac{dt}{\sqrt{{{\left( 2 \right)}^{2}}-{{\left( t \right)}^{2}}}}}={{\sin }^{-1}}\left( \dfrac{t}{2} \right)+C={{\sin }^{-1}}\left( \dfrac{\sin x}{2} \right)+C$, where $C$ is any arbitrary constant. 


10. $\dfrac{\text{si}{{\text{n}}^{\text{8}}}\text{x-co}{{\text{s}}^{\text{8}}}\text{x}}{\text{1-2si}{{\text{n}}^{\text{2}}}\text{xco}{{\text{s}}^{\text{2}}}\text{x}}$.

Ans: The given expression is, $\dfrac{{{\sin }^{8}}x-{{\cos }^{8}}x}{1-2{{\sin }^{2}}x{{\cos }^{2}}x}$. 

$\therefore \dfrac{{{\sin }^{8}}x-{{\cos }^{8}}x}{1-2{{\sin }^{2}}x{{\cos }^{2}}x}=\dfrac{\left( {{\sin }^{4}}x-{{\cos }^{4}}x \right)\left( {{\sin }^{4}}x+{{\cos }^{4}}x \right)}{{{\sin }^{2}}x+{{\cos }^{2}}x-{{\sin }^{2}}x{{\cos }^{2}}x-{{\sin }^{2}}x{{\cos }^{2}}x}$

$=\dfrac{\left( {{\sin }^{2}}x+{{\cos }^{2}}x \right)\left( {{\sin }^{2}}x-{{\cos }^{2}}x \right)\left( {{\sin }^{4}}x+{{\cos }^{4}}x \right)}{{{\sin }^{2}}x\left( 1-{{\cos }^{2}}x \right)+{{\cos }^{2}}x\left( 1-{{\sin }^{2}}x \right)}$

$=\dfrac{-\left( -{{\sin }^{2}}x+{{\cos }^{2}}x \right)\left( {{\sin }^{4}}x+{{\cos }^{4}}x \right)}{{{\sin }^{4}}x+{{\cos }^{4}}x}=-\cos 2x$

Now, integrate the given expression as shown below

$\therefore \int{\dfrac{{{\sin }^{8}}x-{{\cos }^{8}}x}{1-2{{\sin }^{2}}x{{\cos }^{2}}x}dx}=\int{-\cos 2xdx}=\dfrac{-\sin 2x}{2}+C$, where $C$ is any arbitrary constant.

 

11. $\dfrac{\text{1}}{\text{cos}\left( \text{x+a} \right)\text{cos}\left( \text{x+b} \right)}$.

Ans: The given expression is, $\dfrac{1}{\cos \left( x+a \right)\cos \left( x+b \right)}$. 

On multiplying and dividing by $\sin \left( \alpha -\beta  \right)$, the following can be obtained as shown below,

$\dfrac{1}{\sin \left( a-b \right)}\left[ \dfrac{\sin \left( a-b \right)}{\cos \left( x+a \right)\cos \left( x+b \right)} \right]$ 

$\Rightarrow \dfrac{1}{\sin \left( a-b \right)}\left[ \dfrac{\sin \left[ \left( x+a \right)-\left( x+b \right) \right]}{\cos \left( x+a \right)\cos \left( x+b \right)} \right]$

$\Rightarrow \dfrac{1}{\sin \left( a-b \right)}\left[ \dfrac{\sin \left( x+a \right)\cos \left( x+b \right)-\cos \left( x+a \right)\sin \left( x+b \right)}{\cos \left( x+a \right)\cos \left( x+b \right)} \right]$

$\Rightarrow \dfrac{1}{\sin \left( a-b \right)}\left[ \dfrac{\sin (x+a)}{\cos (x+a)}-\dfrac{\sin \left( x+b \right)}{\cos (x+b)} \right]=\dfrac{1}{\sin (a-b)}\left( \tan (x+a)-\tan (x+b) \right)$

$\therefore \int{\dfrac{1}{\cos \left( x+a \right)\cos \left( x+b \right)}dx}=\dfrac{1}{\sin (a-b)}\int{\left( \tan (x+a)-\tan (x+b) \right)dx}$ 

$\Rightarrow \int{\dfrac{1}{\cos \left( x+a \right)\cos \left( x+b \right)}dx}=\dfrac{1}{\sin (a-b)}\left[ \log \left| \dfrac{\cos (x+b)}{\cos (x-a)} \right| \right]+C$, where $C$ is any arbitrary constant.


12. $\dfrac{{{\text{x}}^{\text{3}}}}{\sqrt{\text{1-}{{\text{x}}^{\text{8}}}}}$.

Ans: The given expression is, $\dfrac{{{x}^{3}}}{\sqrt{1-{{x}^{8}}}}$. 

 Now, substitute ${{x}^{4}}=t$ 

$\therefore 4{{x}^{3}}dx=dt$

$\therefore\int{\dfrac{{{x}^{3}}}{\sqrt{1-{{x}^{8}}}}dx}=\dfrac{1}{4}\int{\dfrac{dt}{\sqrt{1-{{\left( t \right)}^{2}}}}}=\dfrac{1}{4}{{\sin }^{-1}}t+C=\dfrac{1}{4}{{\sin }^{-1}}\left( {{x}^{4}} \right)+C$, where $C$ is any arbitrary constant. 


13. $\dfrac{{{\text{e}}^{\text{x}}}}{\left( \text{1+}{{\text{e}}^{\text{x}}} \right)\left( \text{2+}{{\text{e}}^{\text{x}}} \right)}$.

Ans: The given expression is, $\dfrac{{{e}^{x}}}{\left( 1+{{e}^{x}} \right)\left( 2+{{e}^{x}} \right)}$. 

 Now, substitute ${{e}^{x}}=t$ 

$\therefore {{e}^{x}}dx=dt$

$\therefore \int{\dfrac{{{e}^{x}}}{\left( 1+{{e}^{x}} \right)\left( 2+{{e}^{x}} \right)}dx}=\int{\dfrac{dt}{\left( t+1 \right)\left( t+2 \right)}}=\int{\left[ \dfrac{1}{\left( t+1 \right)}-\dfrac{1}{\left( t+2 \right)} \right]dt}$

$\Rightarrow \int{\dfrac{{{e}^{x}}}{\left( 1+{{e}^{x}} \right)\left( 2+{{e}^{x}} \right)}dx}=\log \left| t+1 \right|-\log \left| t+2 \right|+C=\log \left| \dfrac{{{e}^{x}}+1}{{{e}^{x}}+2} \right|+C$, where $C$ is any arbitrary constant. 


14. $\dfrac{\text{1}}{\text{(}{{\text{x}}^{\text{2}}}\text{+1)}\left( {{\text{x}}^{\text{2}}}\text{+4} \right)}$. 

Ans: The given expression is, $\dfrac{1}{({{x}^{2}}+1)\left( {{x}^{2}}+4 \right)}$. 

Now consider it as shown below,

$\therefore \dfrac{1}{({{x}^{2}}+1)\left( {{x}^{2}}+4 \right)}=\dfrac{Ax+B}{\left( {{x}^{2}}+1 \right)}+\dfrac{Cx+D}{\left( {{x}^{2}}+4 \right)}\,\,\,\,\,\,\,\,\,\,\,...\left( 1 \right)$

$\Rightarrow 1=\left( Ax+B \right)\left( {{x}^{2}}+4 \right)+\left( Bx+C \right)\left( {{x}^{2}}+9 \right)$

$\Rightarrow 1=A{{x}^{3}}+4Ax+B{{x}^{2}}+4B+C{{x}^{3}}+Cx+D{{x}^{2}}+D$

On equating the coefficients of ${{x}^{3}},\,{{x}^{2}},\,x$ and constant term, it can be obtained that,  

$A+C=0\,\,\,...\left( 2 \right)$ 

$B+D=0\,\,\,...\left( 3 \right)$

$4A+C=0\,\,\,...\left( 4 \right)$

$4B+D=1\,\,\,...\left( 5 \right)$

And on solving these equations, the values of $A,\,B,\,C,D$ can be obtained as,

$A=0,\,B=\dfrac{1}{3},\,C=0,\,D=\dfrac{1}{3}$ respectively.

Now, from equation $\left( 1 \right)$ it can be clearly obtained that,

$\int{\dfrac{1}{({{x}^{2}}+1)\left( {{x}^{2}}+4 \right)}dx}=\dfrac{1}{3}\int{\left\{ \dfrac{1}{\left( {{x}^{2}}+1 \right)}-\dfrac{1}{\left( {{x}^{2}}+4 \right)}\, \right\}dx}$ 

$=\dfrac{1}{3}{{\tan }^{-1}}x-\dfrac{1}{\left( 3 \right)\left( 2 \right)}{{\tan }^{-1}}\dfrac{x}{2}+C$

$=\dfrac{1}{3}{{\tan }^{-1}}x-\dfrac{1}{6}{{\tan }^{-1}}\dfrac{x}{2}+C$, where $C$ is any arbitrary constant.


15. $\text{co}{{\text{s}}^{\text{3}}}\text{x}{{\text{e}}^{\text{logsinx}}}$.

Ans: The given expression is, ${{\cos }^{3}}x{{e}^{\log \sin x}}$. 

Observe as shown below,

$\therefore {{\cos }^{3}}x{{e}^{\log \sin x}}={{\cos }^{3}}x\sin x$

Now, consider $cosx=t$ 

$\therefore -\sin xdx=dt$

$\therefore \int{{{\cos }^{3}}x{{e}^{\log \sin x}}dx}=\int{{{\cos }^{3}}x\sin xdx}=-\int{{{t}^{3}}dt}=\dfrac{-{{t}^{4}}}{4}+C=\dfrac{-{{\cos }^{4}}x}{4}+C$, where $C$ is any arbitrary constant. 


16. ${{\text{e}}^{\text{3logx}}}{{\left( {{\text{x}}^{\text{4}}}\text{+1} \right)}^{\text{-1}}}$.

Ans: The given expression is, ${{e}^{3\log x}}{{\left( {{x}^{4}}+1 \right)}^{-1}}$. 

Observe as shown below,

$\therefore {{e}^{3\log x}}{{\left( {{x}^{4}}+1 \right)}^{-1}}=\dfrac{{{x}^{3}}}{\left( {{x}^{4}}+1 \right)}$

Now, consider ${{x}^{4}}+1=t$ 

$\therefore 4{{x}^{3}}dx=dt$

$\therefore \int{{{e}^{3\log x}}{{\left( {{x}^{4}}+1 \right)}^{-1}}dx}=\int{\dfrac{{{x}^{3}}}{{{x}^{4}}+1}dx}=\dfrac{1}{4}\int{\dfrac{dt}{t}}=\dfrac{1}{4}\log \left| t \right|+C=\dfrac{\log \left| {{x}^{4}}+1 \right|}{4}+C$, where $C$ is any arbitrary constant. 


17. $\text{f }\!\!'\!\!\text{ }\left( \text{ax+b} \right){{\left[ \text{f(ax+b)} \right]}^{\text{n}}}$.

Ans: The given expression is, $f'\left( ax+b \right){{\left[ f(ax+b) \right]}^{n}}$. 

Now, consider $\left[ f(ax+b) \right]=t$ 

$\therefore af'(ax+b)dx=dt$

$\therefore \int{f'\left( ax+b \right){{\left[ f(ax+b) \right]}^{n}}dx}=\dfrac{1}{a}\int{{{t}^{n}}dt}=\dfrac{{{t}^{n+1}}}{a\left( n+1 \right)}+C=\dfrac{{{\left[ f(ax+b) \right]}^{n+1}}}{a\left( n+1 \right)}+C$, where $C$ is any arbitrary constant. 


18. $\dfrac{\text{1}}{\sqrt{\text{si}{{\text{n}}^{\text{3}}}\text{xsin}\left( \text{x+ }\!\!\alpha\!\!\text{ } \right)}}$.

Ans: The given expression is, $\dfrac{1}{\sqrt{{{\sin }^{3}}x\sin \left( x+\alpha  \right)}}$. 

$\therefore \dfrac{1}{\sqrt{{{\sin }^{3}}x\sin \left( x+\alpha  \right)}}=\dfrac{1}{\sqrt{{{\sin }^{3}}x\sin x\cos \alpha +\cos \alpha \sin x}}$

$=\dfrac{1}{\sqrt{{{\sin }^{4}}x\cos \alpha +{{\sin }^{3}}x\sin \alpha \cos x}}$

$=\dfrac{1}{{{\sin }^{2}}x\sqrt{\cos \alpha +\sin \alpha \cot x}}=\dfrac{\cos e{{c}^{2}}x}{\sqrt{\cos \alpha +\sin \alpha \cot x}}$

Now, substitute $\cos \alpha +\sin \alpha \cot x=t$ 

$\therefore -\cos e{{c}^{2}}x\sin \alpha dx=dt$

$  \therefore \int{\dfrac{1}{\sqrt{{{\sin }^{3}}x\sin \left( x+\alpha  \right)}}dx}=\int{\dfrac{\cos e{{c}^{2}}x}{\sqrt{\cos \alpha +\sin \alpha \cot x}}}=\dfrac{-1}{\sin \alpha }\int{\dfrac{dt}{\sqrt{t}}}=\dfrac{-2\sqrt{t}}{\sin \alpha }+C $  

$  \Rightarrow \int{\dfrac{1}{\sqrt{{{\sin }^{3}}x\sin \left( x+\alpha  \right)}}dx}=\dfrac{-2\sqrt{\cos \alpha +\sin \alpha \cot x}}{\sin \alpha }+C=\dfrac{-2}{\sin \alpha }\sqrt{\cos \alpha +\dfrac{\sin \alpha \cos x}{\sin x}}+C $ 

, where $C$ is an arbitrary constant. 


19. $\sqrt{\dfrac{\text{1-}\sqrt{\text{x}}}{\text{1+}\sqrt{\text{x}}}}$.

Ans: The given expression is, $\sqrt{\dfrac{1-\sqrt{x}}{1+\sqrt{x}}}$. 

Assume, $I=\int{\sqrt{\dfrac{1-\sqrt{x}}{1+\sqrt{x}}}dx}$ 

Now, substitute $x={{\cos }^{2}}\theta $ 

$\therefore dx=-2\sin \theta \cos \theta dt$

$\therefore I=\int{\sqrt{\dfrac{1-\cos \theta }{1+\cos \theta }}\left( -2\sin \theta \cos \theta  \right)d\theta }=-\int{\sqrt{\dfrac{2{{\sin }^{2}}\dfrac{\theta }{2}}{2{{\cos }^{2}}\dfrac{\theta }{2}}}\sin 2\theta d\theta }=-\int{\tan \dfrac{\theta }{2}2\sin \theta \cos \theta d\theta }$

$\Rightarrow I=-2\int{\dfrac{\sin \dfrac{\theta }{2}}{\cos \dfrac{\theta }{2}}2\sin \dfrac{\theta }{2}\cos \dfrac{\theta }{2}\cos \theta d\theta }=-4\int{{{\sin }^{2}}\dfrac{\theta }{2}(2{{\cos }^{2}}\dfrac{\theta }{2}-1)d\theta }$

$\Rightarrow I=-8\int{{{\sin }^{2}}\dfrac{\theta }{2}{{\cos }^{2}}\dfrac{\theta }{2}d\theta }+4\int{{{\sin }^{2}}\dfrac{\theta }{2}d}\theta =2\int{{{\sin }^{2}}\dfrac{\theta }{2}d\theta }+4\int{{{\sin }^{2}}\dfrac{\theta }{2}d}\theta $

$\Rightarrow I=-2\int{\left( \dfrac{1-\cos 2\theta }{2} \right)d\theta }+4\int{\left( \dfrac{1-\cos \theta }{2} \right)d}\theta =-2\left[ \dfrac{\theta }{2}-\dfrac{\sin 2\theta }{2} \right]+4\left[ \dfrac{\theta }{2}-\dfrac{\sin \theta }{2} \right]+C$$\Rightarrow I=-\theta +\dfrac{\sin 2\theta }{2}+2\sin \theta +C=\theta +\sqrt{1-{{\cos }^{2}}\theta }\cos \theta -2\sqrt{1-{{\cos }^{2}}\theta }+C$

$\Rightarrow I={{\cos }^{-1}}\sqrt{x}+\sqrt{x(1-x)}-2\sqrt{1-x}+C=-2\sqrt{1-x}+{{\cos }^{-1}}\sqrt{x}+\sqrt{x-{{x}^{2}}}+C$, where $C$ is any arbitrary constant. 


20. $\dfrac{\text{2+sin2x}}{\text{1+cos2x}}{{\text{e}}^{\text{x}}}$.

Ans: The given expression is, $\dfrac{2+\sin 2x}{1+\cos 2x}{{e}^{x}}$. 

Assume, $I=\int{\dfrac{2+\sin 2x}{1+\cos 2x}{{e}^{x}}dx}$ 

$\Rightarrow I=\int{\dfrac{2+2\sin x\cos x}{2{{\cos }^{2}}x}{{e}^{x}}dx}=\int{\left( \dfrac{1+\sin x\cos x}{{{\cos }^{2}}x} \right){{e}^{x}}dx}=\int{\left( {{\sec }^{2}}x+\tan x \right){{e}^{x}}dx}$

Now, consider $f\left( x \right)=\tan x$ 

$\therefore f'\left( x \right)={{\sec }^{2}}xdx$

$\therefore I=\int{\dfrac{2+\sin 2x}{1+\cos 2x}{{e}^{x}}}dx=\int{\left( f\left( x \right)+f'\left( x \right) \right){{e}^{x}}dx}={{e}^{x}}f\left( x \right)+C={{e}^{x}}\tan x+C$, where $C$ is any arbitrary constant. 


21. $\dfrac{{{\text{x}}^{\text{2}}}\text{+x+1}}{{{\text{(x+1)}}^{\text{2}}}\left( \text{x+2} \right)}$.

Ans: The given expression is, $\dfrac{{{x}^{2}}+x+1}{{{(x+1)}^{2}}\left( x+2 \right)}$. 

Now consider it as shown below,

$\therefore \dfrac{{{x}^{2}}+x+1}{{{(x+1)}^{2}}\left( x+2 \right)}=\dfrac{A}{\left( x+1 \right)}+\dfrac{B}{\left( x+1 \right)}+\dfrac{C}{\left( x+2 \right)}\,\,\,\,\,\,\,\,\,\,\,...\left( 1 \right)$

$\Rightarrow {{x}^{2}}+x+1=A\left( x+1 \right)\left( x+2 \right)+B\left( x+2 \right)+C{{\left( x+1 \right)}^{2}}$

$\Rightarrow {{x}^{2}}+x+1=A\left( {{x}^{2}}+3x+2 \right)+B\left( x+2 \right)+C\left( {{x}^{2}}+2x+1 \right)$

$\Rightarrow {{x}^{2}}+x+1=\left( A+C \right){{x}^{2}}+x\left( 3A+B+2C \right)+\left( 2A+2B+C \right)$

On equating the coefficients of ${{x}^{2}},\,x$ and constant term, it can be obtained that,  

$A+C=1\,\,\,...\left( 2 \right)$ 

$3A+B+2C=1\,\,\,...\left( 3 \right)$

$2A+2B+C=1\,\,\,...\left( 4 \right)$

And on solving these equations, the values of $A,\,B,\,C$ can be obtained as,

$A=-2,\,B=1,\,C=3$ respectively.

Now, from equation $\left( 1 \right)$ it can be clearly obtained that,

$\int{\dfrac{{{x}^{2}}+x+1}{{{(x+1)}^{2}}\left( x+2 \right)}dx}=\int{\left\{ \dfrac{-2}{\left( x+1 \right)}+\dfrac{1}{\left( x+1 \right)}+\dfrac{3}{\left( x+2 \right)} \right\}dx}$ 

$=-2\int{\dfrac{1}{x+1}dx}+\int{\dfrac{1}{{{\left( x+1 \right)}^{2}}}dx}+3\int{\dfrac{1}{x+2}dx}$

$=-2\log \left| x+1 \right|+3\log \left| x+2 \right|-\dfrac{1}{(x+1)}+C$, where $C$ is any arbitrary constant.


22. $\text{ta}{{\text{n}}^{\text{-1}}}\sqrt{\dfrac{\text{1-x}}{\text{1+x}}}$.

Ans: The given expression is, ${{\tan }^{-1}}\sqrt{\dfrac{1-x}{1+x}}$. 

Assume, $I=\int{{{\tan }^{-1}}\sqrt{\dfrac{1-x}{1+x}}dx}$ 

Now, consider $x=\cos \theta $ 

$\therefore dx=-\sin \theta d\theta $

$\therefore I=\int{{{\tan }^{-1}}\sqrt{\dfrac{1-\cos \theta }{1+\cos \theta }}\left( -\sin \theta  \right)d\theta }=-\int{{{\tan }^{-1}}\sqrt{\dfrac{2{{\sin }^{2}}\dfrac{\theta }{2}}{2{{\cos }^{2}}\dfrac{\theta }{2}}}\sin \theta d\theta }=-\int{{{\tan }^{-1}}\tan \dfrac{\theta }{2}\sin \theta d\theta }$ $\Rightarrow I=-\dfrac{1}{2}\int{\theta \sin \theta d\theta }=-\dfrac{1}{2}\left[ \theta \left( -\cos \theta  \right)-\int{1.(-\cos \theta )d\theta } \right]=-\dfrac{1}{2}\left[ \theta \left( \cos \theta  \right)+\sin \theta  \right]$$\Rightarrow I=\dfrac{x}{2}{{\cos }^{-1}}x-\dfrac{1}{2}\sqrt{1-{{x}^{2}}}+C=\dfrac{1}{2}\left( x{{\cos }^{-1}}x-\sqrt{1-{{x}^{2}}} \right)+C$, where $C$ is any arbitrary constant.


23. $\dfrac{\sqrt{{{\text{x}}^{\text{2}}}\text{+1}}\left[ \text{log}\left( {{\text{x}}^{\text{2}}}\text{+1} \right)\text{-2logx} \right]}{{{\text{x}}^{\text{4}}}}$.

Ans: The given expression is, $\dfrac{\sqrt{{{x}^{2}}+1}\left[ \log \left( {{x}^{2}}+1 \right)-2\log x \right]}{{{x}^{4}}}$. 

Assume, $I=\int{\dfrac{\sqrt{{{x}^{2}}+1}\left[ \log \left( {{x}^{2}}+1 \right)-2\log x \right]}{{{x}^{4}}}dx}$ $\Rightarrow I=\dfrac{\sqrt{{{x}^{2}}+1}}{{{x}^{4}}}\left[ \log \left( {{x}^{2}}+1 \right)-\log {{x}^{2}} \right]=\dfrac{\sqrt{{{x}^{2}}+1}}{{{x}^{4}}}\left[ \log \left( \dfrac{{{x}^{2}}+1}{{{x}^{2}}} \right) \right]=\dfrac{1}{{{x}^{3}}}\sqrt{\dfrac{{{x}^{2}}+1}{{{x}^{2}}}}\left[ \log \left( 1+\dfrac{1}{{{x}^{2}}} \right) \right]$Consider $1+\dfrac{1}{{{x}^{2}}}=t$ 

$\therefore \dfrac{-2}{{{x}^{3}}}dx=dt$

Now, integrate the given expression as shown below $\therefore I=\int{\dfrac{\sqrt{{{x}^{2}}+1}\left[ \log \left( {{x}^{2}}+1 \right)-2\log x \right]}{{{x}^{4}}}dx}=\int{\dfrac{1}{{{x}^{3}}}\sqrt{\dfrac{{{x}^{2}}+1}{{{x}^{2}}}}\left[ \log \left( 1+\dfrac{1}{{{x}^{2}}} \right) \right]dx}=\dfrac{-1}{2}\int{{{t}^{\dfrac{1}{2}}}\log tdt}+C$Using integration by parts, it can be obtained that,

$I=\dfrac{-1}{2}\left[ \log t.\int{{{t}^{\dfrac{1}{2}}}dt-\left\{ \left( \dfrac{d}{dt}\log t \right)\int{{{t}^{\dfrac{1}{2}}}dt} \right\}dt} \right]=\dfrac{-1}{2}\left[ \log t.\dfrac{{{t}^{\dfrac{3}{2}}}}{\dfrac{3}{2}}-\int{\dfrac{1}{t}.\dfrac{{{t}^{\dfrac{3}{2}}}}{\dfrac{3}{2}}dt} \right]$ 

$\Rightarrow I=\dfrac{-1}{2}\left[ \dfrac{2}{3}{{t}^{\dfrac{3}{2}}}\log t-\dfrac{2}{3}\int{{{t}^{\dfrac{1}{2}}}dt} \right]=\dfrac{-1}{3}{{t}^{\dfrac{3}{2}}}\log t+\dfrac{2}{9}{{t}^{\dfrac{3}{2}}}=\dfrac{-1}{3}{{t}^{\dfrac{3}{2}}}\left[ \log t-\dfrac{2}{3} \right]$

$\Rightarrow I=\dfrac{-1}{2}\left[ \dfrac{2}{3}{{t}^{\dfrac{3}{2}}}\log t-\dfrac{2}{3}\int{{{t}^{\dfrac{1}{2}}}dt} \right]=\dfrac{-1}{3}{{t}^{\dfrac{3}{2}}}\log t+\dfrac{2}{9}{{t}^{\dfrac{3}{2}}}=\dfrac{-1}{3}{{t}^{\dfrac{3}{2}}}\left[ \log t-\dfrac{2}{3} \right]$

$\Rightarrow I=\dfrac{-1}{2}\left[ 1+\dfrac{1}{{{x}^{2}}} \right]\left( \log \left( 1+\dfrac{1}{{{x}^{2}}} \right)-\dfrac{2}{3} \right)+C$, where $C$ is any arbitrary constant. 


Evaluate the definite integrals in Exercises 24 to 31.

24. $\int_{\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{2}}}^{\text{ }\!\!\pi\!\!\text{ }}{{{\text{e}}^{\text{x}}}\left( \dfrac{\text{1-sinx}}{\text{1-cosx}} \right)\text{dx}}$.

Ans: The given expression is, $\int_{\dfrac{\pi }{2}}^{\pi }{{{e}^{x}}\left( \dfrac{1-\sin x}{1-\cos x} \right)dx}$. 

Assume, $I=\int_{\dfrac{\pi }{2}}^{\pi }{{{e}^{x}}\left( \dfrac{1-\sin x}{1-\cos x} \right)dx}$ 

$\Rightarrow I=\int_{\dfrac{\pi }{2}}^{\pi }{{{e}^{x}}\left( \dfrac{1-2\sin \dfrac{x}{2}\cos \dfrac{x}{2}}{+2{{\sin }^{2}}\dfrac{x}{2}} \right)dx=}\int_{\dfrac{\pi }{2}}^{\pi }{{{e}^{x}}\left( \dfrac{\cos e{{c}^{2}}\dfrac{x}{2}}{2}-\cot \dfrac{x}{2} \right)dx}$

Now, substitute $f\left( x \right)=-\cot \dfrac{x}{2}$ 

$\therefore f'\left( x \right)=-\left( -\dfrac{1}{2}\cos e{{c}^{2}}\dfrac{x}{2} \right)dx=\dfrac{1}{2}\cos e{{c}^{2}}\dfrac{x}{2}dx$

$\therefore I=\int_{\dfrac{\pi }{2}}^{\pi }{{{e}^{x}}\left( f(x)+f'(x) \right)dx}=\left[ {{e}^{x}}f(x) \right]_{\dfrac{\pi }{2}}^{\pi }=\left[ {{e}^{x}}\cot \dfrac{x}{2} \right]_{\dfrac{\pi }{2}}^{\pi }$

$\Rightarrow I=\left[ {{e}^{\pi }}\cot \dfrac{\pi }{2}-{{e}^{\dfrac{\pi }{2}}}\cot \dfrac{\pi }{4} \right]=\left[ 0-{{e}^{\dfrac{\pi }{2}}} \right]=-{{e}^{\dfrac{\pi }{2}}}$


25. $\int_{\text{0}}^{\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{4}}}{\dfrac{\text{sinxcosx}}{\text{si}{{\text{n}}^{\text{4}}}\text{x+co}{{\text{s}}^{\text{4}}}\text{x}}\text{dx}}$.

Ans: The given expression is, $\int_{0}^{\dfrac{\pi }{4}}{\dfrac{\sin x\cos x}{{{\sin }^{4}}x+{{\cos }^{4}}x}dx}$. 

Assume, $I=\int_{0}^{\dfrac{\pi }{4}}{\dfrac{\sin x\cos x}{{{\sin }^{4}}x+{{\cos }^{4}}x}dx}$ 

$\Rightarrow I=\int_{0}^{\dfrac{\pi }{4}}{\dfrac{\dfrac{\sin x\cos x}{{{\cos }^{4}}x}}{\dfrac{{{\sin }^{4}}x+{{\cos }^{4}}x}{{{\cos }^{4}}x}}dx}=\int_{0}^{\dfrac{\pi }{4}}{\dfrac{\tan x{{\sec }^{2}}x}{1+{{\tan }^{4}}x}dx}$

Now, substitute ${{\tan }^{2}}x=t$ 

$\therefore 2\tan x{{\sec }^{2}}xdx=dt$

And also when $x=0,\,t=0$ and when $x=\dfrac{\pi }{4},\,t=1$.

$\therefore I=\dfrac{1}{2}\int_{0}^{1}{\dfrac{dt}{1+{{t}^{2}}}}=\dfrac{1}{2}\left[ {{\tan }^{-1}}t \right]_{0}^{1}=\dfrac{1}{2}\left[ {{\tan }^{-1}}\left( 1 \right)-{{\tan }^{-1}}\left( 0 \right) \right]=\dfrac{1}{2}\left( \dfrac{\pi }{4} \right)=\dfrac{\pi }{8}$


26. $\int_{\text{0}}^{\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{2}}}{\dfrac{\text{co}{{\text{s}}^{\text{2}}}\text{x}}{\text{co}{{\text{s}}^{\text{2}}}\text{x+4si}{{\text{n}}^{\text{2}}}\text{x}}\text{dx}}$.

Ans: The given expression is, $\int_{0}^{\dfrac{\pi }{2}}{\dfrac{{{\cos }^{2}}x}{{{\cos }^{2}}x+4{{\sin }^{2}}x}dx}$. 

Assume, $I=\int_{0}^{\dfrac{\pi }{2}}{\dfrac{{{\cos }^{2}}x}{{{\cos }^{2}}x+4{{\sin }^{2}}x}dx}$ 

$\Rightarrow I=\int_{0}^{\dfrac{\pi }{2}}{\dfrac{{{\cos }^{2}}x}{{{\cos }^{2}}x+4\left( 1-{{\cos }^{2}}x \right)}dx}=\dfrac{-1}{3}\int_{0}^{\dfrac{\pi }{2}}{\dfrac{4-4-3{{\cos }^{2}}x}{-3{{\cos }^{2}}x+4}dx}$

$\Rightarrow I=-\dfrac{1}{3}\int_{0}^{\dfrac{\pi }{2}}{\dfrac{4-3{{\cos }^{2}}x}{4-3{{\cos }^{2}}x}dx}+\dfrac{1}{3}\int_{0}^{\dfrac{\pi }{2}}{\dfrac{4}{4-3{{\cos }^{2}}x}dx}=-\dfrac{1}{3}\int_{0}^{\dfrac{\pi }{2}}{dx}+\dfrac{1}{3}\int_{0}^{\dfrac{\pi }{2}}{\dfrac{4se{{c}^{2}}x}{4{{\sec }^{2}}x-3}dx}$

$\Rightarrow I=\dfrac{-1}{3}\left[ x \right]_{0}^{\dfrac{\pi }{2}}+\dfrac{1}{3}\int_{0}^{\dfrac{\pi }{2}}{\dfrac{4se{{c}^{2}}x}{4\left( 1+{{\tan }^{2}}x \right)-3}dx}=\dfrac{-\pi }{6}+\dfrac{2}{3}\int_{0}^{\dfrac{\pi }{2}}{\dfrac{2se{{c}^{2}}x}{\left( 1+4{{\tan }^{2}}x \right)}dx}\,\,\,...\left( 1 \right)$

Observe, $\int_{0}^{\dfrac{\pi }{2}}{\dfrac{2se{{c}^{2}}x}{\left( 1+4{{\tan }^{2}}x \right)}dx}$

Now, substitute $2\tan x=t$ 

$\therefore 2{{\sec }^{2}}xdx=dt$

And also when $x=0,\,t=0$ and when $x=\dfrac{\pi }{2},\,t=\infty $. 

$\therefore \int_{0}^{\dfrac{\pi }{2}}{\dfrac{2se{{c}^{2}}x}{\left( 1+4{{\tan }^{2}}x \right)}dx}=\int_{0}^{\infty }{\dfrac{dt}{\left( 1+{{t}^{2}} \right)}dx}=\left[ {{\tan }^{-1}}\left( t \right) \right]_{0}^{\infty }=\left[ {{\tan }^{-1}}\left( \infty  \right)-{{\tan }^{-1}}\left( 0 \right) \right]=\dfrac{\pi }{2}$Henceforth from equation $\left( 1 \right)$, it can be obtained that, 

$I=-\dfrac{\pi }{6}+\dfrac{2}{3}\left( \dfrac{\pi }{2} \right)=-\dfrac{\pi }{6}+\dfrac{2\pi }{6}=\dfrac{\pi }{6}$


27. $\int_{\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{6}}}^{\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{2}}}{\dfrac{\text{sinx+cosx}}{\sqrt{\text{sin2x}}}\text{dx}}$.

Ans: The given expression is, $\int_{\dfrac{\pi }{6}}^{\dfrac{\pi }{2}}{\dfrac{\sin x+\cos x}{\sqrt{\sin 2x}}dx}$. 

Assume, $I=\int_{\dfrac{\pi }{6}}^{\dfrac{\pi }{2}}{\dfrac{\sin x+\cos x}{\sqrt{\sin 2x}}dx}$ 

$\Rightarrow I=\int_{0}^{\dfrac{\pi }{2}}{\dfrac{\sin x+\cos x}{\sqrt{-\left( -1+1-2\sin x\cos x \right)}}dx}=\int_{0}^{\dfrac{\pi }{2}}{\dfrac{\sin x+\cos x}{\sqrt{1-\left( {{\sin }^{2}}x+{{\cos }^{2}}x-2\sin x\cos x \right)}}dx}$

$\Rightarrow I=\int_{0}^{\dfrac{\pi }{2}}{\dfrac{\sin x+\cos x}{\sqrt{1-{{\left( \sin x-\cos x \right)}^{2}}}}dx}$

Now, substitute $\left( \sin x-\cos x \right)=t$ 

$\therefore (\sin x+\cos x)dx=dt$

And also when $x=\dfrac{\pi }{6},\,t=\left( \dfrac{1-\sqrt{3}}{2} \right)$ and when $x=\dfrac{\pi }{3},\,t=\left( \dfrac{\sqrt{3}-1}{2} \right)$. 

$\therefore I=\int_{\dfrac{1-\sqrt{3}}{2}}^{\dfrac{\sqrt{3}-1}{2}}{\dfrac{dt}{\sqrt{1-{{t}^{2}}}}}=\int_{-\left( \dfrac{-1+\sqrt{3}}{2} \right)}^{\dfrac{\sqrt{3}-1}{2}}{\dfrac{dt}{\sqrt{1-{{t}^{2}}}}}$

As $\dfrac{1}{\sqrt{1-{{\left( -t \right)}^{2}}}}=\dfrac{1}{\sqrt{1-{{t}^{2}}}}$, it can be thus obtained that $\dfrac{1}{\sqrt{1-{{t}^{2}}}}$ is an even function, 

$\therefore \int_{-a}^{a}{f\left( x \right)dx}=2\int_{0}^{a}{f\left( x \right)dx}$ 

$\therefore I=2\int_{0}^{\dfrac{\sqrt{3}-1}{2}}{\dfrac{dt}{\sqrt{1-{{t}^{2}}}}}=\left[ 2{{\sin }^{-1}}t \right]_{0}^{\dfrac{\sqrt{3}-1}{2}}=2{{\sin }^{-1}}\left( \dfrac{\sqrt{3}-1}{2} \right)$ $x=\dfrac{\pi }{6},\,t=\left( \dfrac{1-\sqrt{3}}{2} \right)$ and when $x=\dfrac{\pi }{3},\,t=\left( \dfrac{\sqrt{3}-1}{2} \right)$. 


28. $\int_{\text{0}}^{\text{1}}{\dfrac{\text{dx}}{\sqrt{\text{1+x}}\text{-}\sqrt{\text{x}}}}$.

Ans: The given expression is, $\int_{0}^{1}{\dfrac{dx}{\sqrt{1+x}-\sqrt{x}}}$. 

Assume, $I=\int_{0}^{1}{\dfrac{dx}{\sqrt{1+x}-\sqrt{x}}}$ 

$\Rightarrow I=\int_{0}^{1}{\dfrac{1}{\sqrt{1+x}-\sqrt{x}}\times \dfrac{\sqrt{1+x}+\sqrt{x}}{\sqrt{1+x}+\sqrt{x}}dx}=\int_{0}^{1}{\dfrac{\sqrt{1+x}+\sqrt{x}}{1+x-x}dx}$

$\Rightarrow I=\int_{0}^{1}{\sqrt{1+x}dx}+\int_{0}^{1}{\sqrt{x}dx}=\dfrac{2}{3}\left[ {{\left( 1+x \right)}^{\dfrac{2}{3}}} \right]_{0}^{1}+\dfrac{2}{3}\left[ {{\left( x \right)}^{\dfrac{3}{2}}} \right]_{0}^{1}=\dfrac{2}{3}\left[ {{\left( 2 \right)}^{\dfrac{2}{3}}}-1 \right]+\dfrac{2}{3}=\dfrac{4\sqrt{2}}{3}$


29. $\int_{\text{0}}^{\dfrac{\text{ }\!\!\pi\!\!\text{}}{\text{4}}}{\dfrac{\text{sinx+cosx}}{\text{9+16sin2x}}\text{dx}}$.

Ans: The given expression is, $\int_{0}^{\dfrac{\pi }{4}}{\dfrac{\sin x+\cos x}{9+16\sin 2x}dx}$. 

Assume, $I=\int_{0}^{\dfrac{\pi }{4}}{\dfrac{\sin x+\cos x}{9+16\sin 2x}dx}$ 

Now, substitute $\sin x-\cos x=t$ 

$\therefore \left( \cos x+\sin x \right)dx=dt$

And also when $x=0,\,t=-1$ and when $x=\dfrac{\pi }{4},\,t=0$.

$\therefore {{\left( \sin x-\cos x \right)}^{2}}={{t}^{2}}$

$\Rightarrow 1-2\sin x\cos x={{t}^{2}}$

$\Rightarrow 1-\sin 2x={{t}^{2}}$

$\Rightarrow \sin 2x=1-{{t}^{2}}$

$\therefore I=\int_{-1}^{0}{\dfrac{dt}{9+16\left( 1-{{t}^{2}} \right)}}=\int_{-1}^{0}{\dfrac{dt}{25-16{{t}^{2}}}}=\int_{-1}^{0}{\dfrac{dt}{{{\left( 5 \right)}^{2}}-{{\left( 4t \right)}^{2}}}}$

$\Rightarrow I=\dfrac{1}{4}\left[ \dfrac{1}{2\left( 5 \right)}\log \left| \dfrac{5+4t}{5-4t} \right| \right]_{-1}^{0}=\dfrac{1}{40}\left[ \log \left| 1 \right|-\log \left| \dfrac{1}{9} \right| \right]=\dfrac{1}{40}\log \left| 9 \right|$


30. $\int_{\text{0}}^{\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{2}}}{\text{sin2xta}{{\text{n}}^{\text{-1}}}\left( \text{sinx} \right)\text{dx}}$.

Ans: The given expression is, $\int_{0}^{\dfrac{\pi }{2}}{\sin 2x{{\tan }^{-1}}\left( \sin x \right)dx}$. 

Assume, $I=\int_{0}^{\dfrac{\pi }{2}}{\sin 2x{{\tan }^{-1}}\left( \sin x \right)dx}$ 

$\Rightarrow I=\int_{0}^{\dfrac{\pi }{2}}{\sin 2x{{\tan }^{-1}}\left( \sin x \right)dx}=\int_{0}^{\dfrac{\pi }{2}}{2\sin x\cos x{{\tan }^{-1}}\left( \sin x \right)dx}$

Now, substitute $\sin x=t$ 

$\therefore \cos xdx=dt$

And also when $x=0,\,t=0$ and when $x=\dfrac{\pi }{2},\,t=1$. 

$\therefore I=2\int_{0}^{1}{t{{\tan }^{-1}}\left( t \right)dt}$

Observe, $\int{t{{\tan }^{-1}}\left( t \right)dt}$

$\therefore \int{t{{\tan }^{-1}}\left( t \right)dt}={{\tan }^{-1}}t\int{tdt}-\int{\left\{ \dfrac{d\left( {{\tan }^{-1}}t \right)}{dt}\int{tdt} \right\}dt}={{\tan }^{-1}}t.\dfrac{{{t}^{2}}}{2}-\int{\dfrac{1}{1+{{t}^{2}}}.\dfrac{{{t}^{2}}}{2}dt}$

$=\dfrac{{{t}^{2}}{{\tan }^{-1}}t}{2}-\int{\dfrac{{{t}^{2}}+1-1}{1+{{t}^{2}}}dt}=\dfrac{{{t}^{2}}{{\tan }^{-1}}t}{2}-\int{1dt}+\int{\dfrac{1}{1+{{t}^{2}}}dt}=\dfrac{{{t}^{2}}{{\tan }^{-1}}t}{2}-\dfrac{1}{2}t+\dfrac{1}{2}{{\tan }^{-1}}t$

$\therefore \int_{0}^{1}{t.{{\tan }^{-1}}tdt}=\left[ \dfrac{{{t}^{2}}{{\tan }^{-1}}t}{2}-\dfrac{1}{2}t+\dfrac{1}{2}{{\tan }^{-1}}t \right]_{0}^{1}=\dfrac{1}{2}\left[ \dfrac{\pi }{4}-1+\dfrac{\pi }{4} \right]=\dfrac{\pi }{4}-\dfrac{1}{2}$

Henceforth from equation $\left( 1 \right)$, it can be obtained that, 

$I=2\left[ \dfrac{\pi }{2}-\dfrac{1}{2} \right]=\dfrac{\pi }{2}-1$


31. $\int_{\text{1}}^{\text{4}}{\left[ \left| \text{x-1} \right|\text{+}\left| \text{x-2} \right|\text{+}\left| \text{x-3} \right| \right]\text{dx}}$.

Ans: The given expression is, $\int_{1}^{4}{\left[ \left| x-1 \right|+\left| x-2 \right|+\left| x-3 \right| \right]dx}$. 

Assume, $\int_{1}^{4}{\left[ \left| x-1 \right|+\left| x-2 \right|+\left| x-3 \right| \right]dx}$ 

$\Rightarrow I=\int_{1}^{4}{\left| x-1 \right|dx}+\int_{1}^{4}{\left| x-2 \right|dx}+\int_{1}^{4}{\left| x-3 \right|dx}$

$\therefore I={{I}_{1}}+{{I}_{2}}+{{I}_{3}}\,\,\,...\left( 1 \right)$

where, ${{I}_{1}}=\int_{1}^{4}{\left| x-1 \right|dx},\,{{I}_{2}}=\int_{1}^{4}{\left| x-2 \right|dx},\,{{I}_{3}}=+\int_{1}^{4}{\left| x-3 \right|dx}$ 

Now, consider, ${{I}_{1}}=\int_{1}^{4}{\left| x-1 \right|dx}$, where $\left( x-1 \right)\ge 0\,\forall \,1\le x\le 4$  $\therefore {{I}_{1}}=\int_{1}^{4}{\left( x-1 \right)dx}=\left[ \dfrac{{{x}^{2}}}{2}-x \right]_{1}^{4}=\left[ 8-4-\dfrac{1}{2}+1 \right]=\dfrac{9}{2}\,\,\,...\left( 2 \right)\,$

Again, consider, ${{I}_{2}}=\int_{1}^{4}{\left| x-2 \right|dx}$, where $\left( x-2 \right)\ge 0\,\forall \,2\le x\le 4$ and $\left( x-2 \right)\le 0\,\forall \,1\le x\le 2$.

$\therefore {{I}_{2}}=\int_{1}^{2}{\left( 2-x \right)dx}+\int_{2}^{4}{\left( x-2 \right)dx}=\left[ 2x-\dfrac{{{x}^{2}}}{2} \right]_{1}^{4}+\left[ \dfrac{{{x}^{2}}}{2}-2x \right]_{2}^{4}$

$\Rightarrow {{I}_{2}}=\left[ 4-2-2+\dfrac{1}{2} \right]+\left[ 8-8-2+4 \right]=\dfrac{1}{2}+2=\dfrac{5}{2}\,\,\,...\left( 3 \right)\,$

Also, consider, ${{I}_{3}}=\int_{1}^{4}{\left| x-3 \right|dx}$, where $\left( x-3 \right)\ge 0\,\forall \,3\le x\le 4$ and $\left( x-3 \right)\le 0\,\forall \,1\le x\le 3$.

$\therefore {{I}_{3}}=\int_{1}^{3}{\left( 3-x \right)dx}+\int_{3}^{4}{\left( x-3 \right)dx}=\left[ 3-\dfrac{{{x}^{2}}}{2} \right]_{1}^{3}+\left[ \dfrac{{{x}^{2}}}{2}-3x \right]_{3}^{4}$

$\Rightarrow {{I}_{3}}=\left[ 9-\dfrac{9}{2}-3+\dfrac{1}{2} \right]+\left[ 8-12-\dfrac{9}{2}+9 \right]=2+\dfrac{1}{2}=\dfrac{5}{2}\,\,\,...\left( 4 \right)\,$

Now, from equations $\left( 1 \right)$, $\left( 2 \right)$, $\left( 3 \right)$ and $\left( 4 \right)$ it can be obtained that, 

$I=\dfrac{9}{2}+\dfrac{5}{2}+\dfrac{5}{2}=\dfrac{19}{2}$

$\Rightarrow 2I=\pi \int_{0}^{\pi }{\dfrac{\sin x+1-1}{1+\sin x}dx}=\pi \int_{0}^{\pi }{dx}-\pi \int_{0}^{\pi }{\dfrac{1}{1+\sin x}dx}=\pi \left[ x \right]_{0}^{\pi }-\pi \int_{0}^{\pi }{\dfrac{1-\sin x}{{{\cos }^{2}}x}dx}$

$\Rightarrow 2I=\pi \left[ x \right]_{0}^{\pi }-\pi \int_{0}^{\pi }{\left( {{\sec }^{2}}x-\tan x\sec x \right)dx}={{\pi }^{2}}-\pi \left[ \tan x-\sec x \right]_{0}^{\pi }$

$\Rightarrow 2I={{\pi }^{2}}-\pi \left[ 0-\left( -1 \right)-0+1 \right]={{\pi }^{2}}-2\pi $

$\Rightarrow I=\dfrac{\pi \left( \pi -2 \right)}{2}$


Prove the following (Exercises 32 to 37)

32. $\int_{\text{1}}^{\text{3}}{\dfrac{\text{dx}}{{{\text{x}}^{\text{2}}}\left( \text{x+1} \right)}}\text{=}\dfrac{\text{2}}{\text{3}}\text{+log}\dfrac{\text{2}}{\text{3}}$.

Ans: The given equation is, $\int_{1}^{3}{\dfrac{dx}{{{x}^{2}}\left( x+1 \right)}}=\dfrac{2}{3}+\log \dfrac{2}{3}$. 

Assume, $\int_{1}^{3}{\dfrac{dx}{{{x}^{2}}\left( x+1 \right)}}$ 

Now consider it as shown below,

$\therefore \dfrac{1}{{{x}^{2}}\left( x+1 \right)}=\dfrac{A}{x}+\dfrac{B}{{{x}^{2}}}+\dfrac{C}{\left( x+1 \right)}\,\,\,\,\,\,\,\,\,\,\,...\left( 1 \right)$

$\Rightarrow 1=Ax\left( x+1 \right)+B\left( x+1 \right)+C\left( {{x}^{2}} \right)$

$\Rightarrow 1=A{{x}^{2}}+Ax+Bx+B+C{{x}^{2}}$

On equating the coefficients of ${{x}^{2}},\,x$ and constant term, it can be obtained that,  

$A+C=0\,\,\,...\left( 2 \right)$ 

$A+B=0\,\,\,...\left( 3 \right)$

$B=1\,\,\,...\left( 4 \right)$

And on solving these equations, the values of $A,\,B,\,C$ can be obtained as,

$A=-1,\,B=1,\,C=1$ respectively.

Now, from equation $\left( 1 \right)$ it can be clearly obtained that,

$I=\int_{1}^{3}{\dfrac{dx}{{{x}^{2}}\left( x+1 \right)}}=\int_{1}^{3}{\left\{ \dfrac{-1}{x}+\dfrac{1}{{{x}^{2}}}+\dfrac{1}{\left( x+1 \right)} \right\}dx}$ 

$\Rightarrow I=\left[ -\log x-\dfrac{1}{x}+\log \left( x+1 \right) \right]_{1}^{3}=\left[ \log \left( \dfrac{x+1}{x} \right)-\dfrac{1}{x} \right]_{1}^{3}=\log \left( \dfrac{4}{3} \right)-\dfrac{1}{3}-\log \left( 2 \right)+1$

$\Rightarrow I=\log 4-\log 3-\log 2+\dfrac{2}{3}=\log 2-\log 3+\dfrac{2}{3}=\log \left( \dfrac{2}{3} \right)+\dfrac{2}{3}$

Henceforth, it can be proved.


33. $\int_{\text{0}}^{\text{4}}{\text{x}{{\text{e}}^{\text{x}}}\text{dx}}\text{=1}$.

Ans: The given equation is, $\int_{0}^{4}{x{{e}^{x}}dx}=1$. 

Assume, $I=\int_{0}^{4}{x{{e}^{x}}dx}$ 

Using integration by parts, it can be obtained that,

$I=x\int_{0}^{4}{{{e}^{x}}dx}-\int_{0}^{4}{\left\{ \left( \dfrac{d\left( x \right)}{dx} \right)\int{{{e}^{x}}} \right\}}=\left[ x{{e}^{x}} \right]_{0}^{1}-\left[ {{e}^{x}} \right]_{0}^{1}=e-e+1=1$ 

Henceforth, it can be proved.


34. $\int_{\text{1}}^{\text{-1}}{{{\text{x}}^{\text{17}}}\text{co}{{\text{s}}^{\text{4}}}\text{xdx}}\text{=0}$.

Ans: The given equation is, $\int_{1}^{-1}{{{x}^{17}}{{\cos }^{4}}xdx}=0$. 

Assume, $I=\int_{1}^{-1}{{{x}^{17}}{{\cos }^{4}}xdx}$ 

Now, consider $f(x)={{x}^{17}}{{\cos }^{4}}x$ 

$\therefore f\left( -x \right)={{\left( -x \right)}^{17}}{{\cos }^{4}}\left( -x \right)=-{{x}^{17}}{{\cos }^{4}}x=-f\left( x \right)$

$\Rightarrow f\left( x \right)$ is an odd function and henceforth it is known to us that when $f(x)$ is an odd function, then $\int_{-a}^{a}{f\left( x \right)dx}=0$. 

$\therefore I=\int_{1}^{-1}{{{x}^{17}}{{\cos }^{4}}xdx}=0$

Henceforth, it can be proved.


35. $\int_{\text{0}}^{\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{2}}}{\text{si}{{\text{n}}^{\text{3}}}\text{xdx}}\text{=}\dfrac{\text{2}}{\text{3}}$.

Ans: The given equation is, $\int_{0}^{\dfrac{\pi }{2}}{{{\sin }^{3}}xdx}=\dfrac{2}{3}$. 

Assume, $I=\int_{0}^{\dfrac{\pi }{2}}{{{\sin }^{3}}xdx}$ 

$\Rightarrow I=\int_{0}^{\dfrac{\pi }{2}}{{{\sin }^{2}}x\sin xdx}=\int_{0}^{\dfrac{\pi }{2}}{\left( 1-{{\cos }^{2}}x \right)\sin xdx}=\int_{0}^{\dfrac{\pi }{2}}{\sin xdx}-\int_{0}^{\dfrac{\pi }{2}}{{{\cos }^{2}}x\sin xdx}$

$\Rightarrow I=\left[ -\cos x \right]_{0}^{\dfrac{\pi }{2}}+\left[ \dfrac{co{{s}^{3}}x}{3} \right]_{0}^{\dfrac{\pi }{2}}=1-\dfrac{1}{3}=\dfrac{2}{3}$

Henceforth, it can be proved.


36. $\int_{\text{0}}^{\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{4}}}{\text{2ta}{{\text{n}}^{\text{3}}}\text{xdx}}\text{=1-log2}$.

Ans: The given equation is, $\int_{0}^{\dfrac{\pi }{4}}{2{{\tan }^{3}}xdx}=1-\log 2$. 

Assume, $\int_{0}^{\dfrac{\pi }{4}}{2{{\tan }^{3}}xdx}$ 

$\Rightarrow I=\int_{0}^{\dfrac{\pi }{4}}{2{{\tan }^{2}}x\tan xdx}=\int_{0}^{\dfrac{\pi }{4}}{\left( 1-{{\sec }^{2}}x \right)\tan xdx}=\int_{0}^{\dfrac{\pi }{4}}{\tan xdx}-\int_{0}^{\dfrac{\pi }{4}}{{{\sec }^{2}}x\tan xdx}$

$\Rightarrow I=2\left[ \dfrac{{{\tan }^{2}}x}{2} \right]_{0}^{\dfrac{\pi }{4}}+2\left[ \log \cos x \right]_{0}^{\dfrac{\pi }{4}}=1+2\left[ \log \cos \dfrac{\pi }{4}-\log \cos 0 \right]_{0}^{\dfrac{\pi }{2}}=1-\log 2-\log 1$

$\Rightarrow I=1-\log 2$

Henceforth, it can be proved.


37. $\int_{\text{0}}^{\text{1}}{\text{si}{{\text{n}}^{\text{-1}}}\text{xdx}}\text{=}\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{2}}\text{-1}$.

Ans: The given equation is, $\int_{0}^{1}{{{\sin }^{-1}}xdx}=\dfrac{\pi }{2}-1$. 

Assume, $I=\int_{0}^{1}{{{\sin }^{-1}}xdx}$ 

$\Rightarrow I=\int_{0}^{1}{{{\sin }^{-1}}x.1dx}$

Using integration by parts, it can be obtained that,

$I=\left[ {{\sin }^{-1}}x.x \right]_{0}^{1}-\int_{0}^{1}{\dfrac{1}{\sqrt{1-{{x}^{2}}}}xdx}=\left[ x{{\sin }^{-1}}x \right]_{0}^{1}+\dfrac{1}{2}\int_{0}^{1}{\dfrac{\left( -2x \right)}{\sqrt{1-{{x}^{2}}}}dx}$ 

Now, substitute $1-{{x}^{2}}=t$ 

$\therefore \left( -2x \right)dx=dt$

And also when $x=0,\,t=1$ and when $x=1,\,t=0$.

$\therefore I=\left[ x{{\sin }^{-1}}x \right]_{0}^{1}+\dfrac{1}{2}\int_{0}^{1}{\dfrac{dt}{\sqrt{t}}}=\left[ x{{\sin }^{-1}}x \right]_{0}^{1}+\dfrac{1}{2}\left[ 2\sqrt{t} \right]_{1}^{0}={{\sin }^{-1}}1-\sqrt{1}=\dfrac{\pi }{2}-1$ Henceforth, it can be clearly proved.


Choose the correct answers in Exercises 38 to 40

38. $\int{\dfrac{\text{dx}}{{{\text{e}}^{\text{x}}}\text{+}{{\text{e}}^{\text{-x}}}}}$ is equal to

A. $\text{ta}{{\text{n}}^{\text{-1}}}\left( {{\text{e}}^{\text{x}}} \right)\text{+C}$

B. $\text{ta}{{\text{n}}^{\text{-1}}}\left( {{\text{e}}^{\text{-x}}} \right)\text{+C}$ 

C. $\text{log}\left( {{\text{e}}^{\text{x}}}\text{-}{{\text{e}}^{\text{-x}}} \right)\text{+C}$ 

D. $\text{log}\left( {{\text{e}}^{\text{x}}}\text{+}{{\text{e}}^{\text{-x}}} \right)\text{+C}$ 

Ans: The given expression is, $\int{\dfrac{dx}{{{e}^{x}}+{{e}^{-x}}}}$. 

Assume, $I=\int{\dfrac{dx}{{{e}^{x}}+{{e}^{-x}}}}$ 

Now, consider ${{e}^{x}}=t$ 

$\therefore {{e}^{x}}dx=dt$

$\therefore I=\int{\dfrac{dx}{{{e}^{x}}+{{e}^{-x}}}}=\int{\dfrac{1}{1+{{t}^{2}}}dt}=\int{{{\tan }^{-1}}\operatorname{t}dt}+C$, where $C$ is any arbitrary constant.

Hence, the correct answer is option (A). 


39. $\int{\dfrac{\text{cos2x}}{{{\left( \text{sinx+cosx} \right)}^{\text{2}}}}\text{dx}}$ is

A. $\dfrac{\text{-1}}{\text{sinx+cosx}}\text{+C}$

B. $\text{log}\left| \text{sinx+cosx} \right|\text{+C}$ 

C. $\text{log}\left| \text{sinx-cosx} \right|\text{+C}$ 

D. $\dfrac{\text{1}}{{{\left( \text{sinx+cosx} \right)}^{\text{2}}}}\text{+C}$ 

Ans: The given expression is, $\int{\dfrac{\cos 2x}{{{\left( \sin x+\cos x \right)}^{2}}}dx}$. 

Assume, $I=\int{\dfrac{\cos 2x}{{{\left( \sin x+\cos x \right)}^{2}}}dx}$ 

$\Rightarrow I=\int{\dfrac{(\sin x+\cos x)(\cos x-\sin x)}{{{\left( \sin x+\cos x \right)}^{2}}}dx}=\int{\dfrac{(\cos x-\sin x)}{\left( \sin x+\cos x \right)}dx}$

Now, substitute $\left( \sin x+\cos x \right)=t$ 

$\therefore (\cos x-\sin x)dx=dt$

$\therefore I=I=\int{\dfrac{1}{t}dt=\log \left| t \right|+C=\log \left| \cos x+\sin x \right|+C}$, where $C$ is any arbitrary constant.

Hence, the correct answer is option (B). 


40. If $\text{f}\left( \text{a+b-x} \right)\text{=f}\left( \text{x} \right)\text{,}\,$then $\int_{\text{a}}^{\text{b}}{\text{xf}\left( \text{x} \right)\text{dx}}$ is equal to

A. $\dfrac{\text{a+b}}{\text{2}}\int_{\text{a}}^{\text{b}}{\text{f}\left( \text{b-x} \right)\text{dx}}$

B. $\dfrac{\text{a+b}}{\text{2}}\int_{\text{a}}^{\text{b}}{\text{f}\left( \text{b+x} \right)\text{dx}}$ 

C. $\dfrac{\text{b-a}}{\text{2}}\int_{\text{a}}^{\text{b}}{\text{f}\left( \text{x} \right)\text{dx}}$ 

D. $\dfrac{\text{a+b}}{\text{2}}\int_{\text{a}}^{\text{b}}{\text{f}\left( \text{x} \right)\text{dx}}$ 

Ans: Assume, $I=\int_{a}^{b}{xf\left( x \right)dx}\,\,\,...\left( 1 \right)$ 

$\Rightarrow I=\int_{a}^{b}{\left( a+b-x \right)f\left( a+b-x \right)dx}\,\,\,\left[ \because \int_{a}^{b}{f\left( x \right)dx}=\int_{a}^{b}{f\left( a+b-x \right)dx} \right]$

$\Rightarrow I=\int_{a}^{b}{\left( a+b-x \right)f\left( x \right)dx}=\left( a+b \right)\int_{a}^{b}{f\left( x \right)dx}-I$   using $\left( 1 \right)$

$\Rightarrow 2I=\left( a+b \right)\int_{a}^{b}{f\left( x \right)dx}$

$\Rightarrow I=\dfrac{\left( a+b \right)}{2}\int_{a}^{b}{f\left( x \right)dx}$

Hence, the correct answer is option (D). 


Conclusion

The Class 12 Maths Chapter 7 Miscellaneous Exercise Solutions are important for understanding various concepts thoroughly. It contains a variety of problems that necessitate the use of multiple formulas and techniques. It's crucial to concentrate on understanding the fundamental principles behind each question rather than simply memorizing solutions. Keep in mind the importance of understanding the theory behind each concept, consistent practice, and referencing solved examples to effectively master this exercise.


Class 12 Maths Chapter 7: Exercises Breakdown

Exercise

Number of Questions

Exercise 7.1

22 Questions and Solutions

Exercise 7.2

39 Questions and Solutions

Exercise 7.3

24 Questions and Solutions

Exercise 7.4

25 Questions and Solutions

Exercise 7.5

23 Questions and Solutions

Exercise 7.6

24 Questions and Solutions

Exercise 7.7

11 Questions and Solutions

Exercise 7.8

22 Questions and Solutions

Exercise 7.9

10 Questions and Solutions

Exercise 7.10

21 Questions and Solutions


CBSE Class 12 Maths Chapter 7 Other Study Materials


Chapter-Specific NCERT Solutions for Class 12 Maths

Given below are the chapter-wise NCERT Solutions for Class 12 Maths. Go through these chapter-wise solutions to be thoroughly familiar with the concepts.


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FAQs on NCERT Solutions for Class 12 Maths Chapter 7 Integrals Miscellaneous Exercise

1. Where can I find step-by-step NCERT Solutions for the Miscellaneous Exercise of Class 12 Maths Chapter 7?

Vedantu provides detailed, step-by-step NCERT Solutions for all questions in the Miscellaneous Exercise of Class 12 Maths Chapter 7 (Integrals). These solutions are prepared by expert teachers and are fully aligned with the CBSE 2025-26 syllabus, explaining the correct method and formulas for each problem.

2. How many questions are in the Miscellaneous Exercise for Integrals, and what is their structure?

The Miscellaneous Exercise for Chapter 7, Integrals, contains 44 questions in the latest NCERT textbook. The solutions cover a structured format designed for comprehensive revision:

  • Indefinite Integrals: Questions 1 to 24
  • Definite Integrals: Questions 25 to 33 and 41 to 44
  • Proof-Based Questions: Questions 34 to 37
  • Multiple Choice Questions (MCQs): Questions 38 to 40
This mix ensures that all concepts from the chapter are tested.

3. What is the correct method to solve Question 1 of the Miscellaneous Exercise, which is ∫ dx / (x - x³)?

To solve ∫ dx / (x - x³), the correct method is integration by partial fractions. The steps are as follows:

  • First, factorise the denominator completely: x(1 - x²) = x(1 - x)(1 + x).
  • Next, express the integrand 1 / [x(1 - x)(1 + x)] in the form A/x + B/(1-x) + C/(1+x).
  • Determine the values of the constants A, B, and C by comparing coefficients.
  • Finally, integrate the resulting simpler rational functions to arrive at the solution: (1/2)log|x² / (1 - x²)| + C.

4. Why are the questions in the Miscellaneous Exercise of Chapter 7 often considered more difficult than those in other exercises?

The questions in the Miscellaneous Exercise are generally more challenging because they are not grouped by a single technique. They require students to first identify the appropriate integration method (like substitution, by parts, partial fractions) or even a combination of methods. This tests a deeper, more holistic understanding of the entire chapter rather than just the application of one specific formula, making it excellent preparation for board exams.

5. How do I solve definite integral problems involving modulus functions, like Question 31, ∫(|x-1|+|x-2|+|x-3|) dx from 1 to 4?

To solve definite integrals with modulus functions, you must split the integral's limits at the points where the expressions inside the modulus become zero. For this problem, the steps are:

  • Identify the critical points x=1, x=2, and x=3. Split the integral at x=2 and x=3 within the given range [1, 4].
  • Rewrite the integral as a sum: ∫(from 1 to 2) + ∫(from 2 to 3) + ∫(from 3 to 4).
  • In each interval, remove the modulus by determining the sign of the functions. For instance, in [1, 2], |x-2| becomes -(x-2).
  • Evaluate each simple integral separately and add the results to get the final answer.

6. In Question 20, ∫ eˣ(2 + sin2x) / (1 + cos2x) dx, why is the property ∫eˣ[f(x) + f'(x)]dx used for the solution?

The property ∫eˣ[f(x) + f'(x)]dx = eˣf(x) + C is used because the integrand can be simplified into this special form, which makes the solution direct.

  • First, use trigonometric half-angle identities: 1 + cos2x = 2cos²x and sin2x = 2sinxcosx.
  • The integrand simplifies to eˣ[(2 + 2sinxcosx) / (2cos²x)], which equals eˣ(sec²x + tanx).
  • If we choose f(x) = tanx, its derivative is f'(x) = sec²x.
  • The integral perfectly matches the form ∫eˣ[f(x) + f'(x)]dx, so the solution is simply eˣtanx + C. This method is significantly faster than others.

7. What is the key property of definite integrals needed to solve Question 40?

The key property of definite integrals required for Question 40, where f(a+b-x) = f(x), is P4: ∫(from a to b) g(x) dx = ∫(from a to b) g(a+b-x) dx. By applying this property to the integral I = ∫(from a to b) x f(x) dx, and using the given condition, you can create a second equation. Adding the two equations for I eliminates the 'x' term and leads to the correct answer: ((a+b)/2) ∫(from a to b) f(x) dx.

8. Why is using a substitution like x = a/t suggested for a problem like Question 3, ∫ dx / (x√(ax - x²))?

The substitution x = a/t is a strategic choice specifically designed to simplify the complex radical expression √(ax - x²). When you substitute x = a/t and the corresponding dx = (-a/t²)dt, the terms inside the square root transform into a much simpler form containing √(t-1). This new integral is a standard one that can be solved easily. Such non-obvious substitutions are a key problem-solving skill tested in the Miscellaneous Exercise.

9. What is the correct approach for integrals with inverse trigonometric functions, like Question 22, ∫ tan⁻¹√((1-x)/(1+x)) dx?

The most effective approach is to use a trigonometric substitution that simplifies the expression inside the inverse function.

  • Let x = cosθ. This choice is ideal because the expressions 1-cosθ and 1+cosθ have standard half-angle formulas.
  • The term √((1-cosθ)/(1+cosθ)) simplifies to tan(θ/2).
  • The integral becomes ∫ tan⁻¹(tan(θ/2)) (-sinθ dθ), which simplifies to -∫(θ/2)(sinθ) dθ.
  • This resulting integral can be solved using integration by parts. Finally, substitute back θ = cos⁻¹x to express the answer in terms of x.

10. How are the solutions for the proof-based definite integrals (Questions 32-37) structured in the NCERT solutions?

The solutions for proof-based questions are structured to show a clear, logical progression from the Left-Hand Side (LHS) to the Right-Hand Side (RHS). The standard steps are:

  • Start with the LHS: Clearly state the integral that needs to be evaluated.
  • Apply Integration Techniques: Use the most suitable method (e.g., partial fractions for Q32, integration by parts for Q33) to find the antiderivative.
  • Apply the Limits: As per the Fundamental Theorem of Calculus, substitute the upper and lower limits of integration.
  • Simplify to match RHS: Perform the necessary arithmetic and algebraic simplification until the expression matches the required RHS.