NCERT Solutions For Class 12 Maths Miscellaneous Exercise Chapter 7 Integrals - 2025-26

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

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.