CBSE Class 12 Mathematics Chapter 7 Integrals – NCERT Solutions Exercise 7.6 [2025-26]

1. $x\sin x$

Ans:  Let $I=\int{x}\sin xdx$

Consider $\text{u}=\text{x}$ and $\text{v}=\sin \text{x}$and integrating by parts, to obtain  

$I=\int{x}\sin xdx-\int{\left\{ \left( \dfrac{d}{dx}x \right)\int{\sin }xdx \right\}}dx$

$=x(-\cos x)-\int{1}.(-\cos x)dx$

$=-x\cos x+\sin x+C$

2. $x\sin 3x$

Ans: Let $\text{I}=\int{x}\sin 3xdx$

Consider $\text{u}=\text{x}$ and $\text{v}=\sin 3\text{x}$ and integrating by parts, to obtain  

$I=x\int{\sin }3xdx-\int{\left\{ \left( \dfrac{d}{dx}x \right)\int{\sin }3xdx \right\}}$

$=x\left( \dfrac{-\cos 3x}{3} \right)-\int{1}\cdot \left( \dfrac{-\cos 3x}{3} \right)dx$

$=\dfrac{-x\cos 3x}{3}+\dfrac{1}{3}\int{\cos }3xdx=\dfrac{-x\cos 3x}{3}+\dfrac{1}{9}\sin 3x+C$

3. ${{x}^{2}}{{e}^{x}}$

Ans:  Let $I=\int{{{x}^{2}}}{{e}^{x}}dx$

Consider  $\text{u}={{\text{x}}^{2}}\,\,and\,\,\text{v}={{\text{e}}^{x}}$

$I={{x}^{2}}\int{{{e}^{x}}}dx-\int{\left\{ \left( \dfrac{d}{dx}{{x}^{2}} \right)\int{{{e}^{x}}}dx \right\}}dx$

$={{x}^{2}}{{e}^{x}}-\int{2}x-{{e}^{x}}dx$

$={{x}^{2}}{{e}^{x}}-2\int{x}\cdot {{e}^{x}}dx$

Again using integration by parts, to obtain

$={{x}^{2}}{{e}^{x}}-2\left[ x\cdot \int{{{e}^{x}}}dx-\int{\left\{ \left( \dfrac{d}{dx}{{x}^{2}} \right)\int{{{e}^{x}}}dx \right\}}dx \right]$

$={{x}^{2}}{{e}^{x}}-2\left[ x{{e}^{x}}-\int{{{e}^{x}}}dx \right]$

Simplifying,

$={{x}^{2}}{{e}^{x}}-2\left[ x{{e}^{x}}-{{e}^{x}} \right]$

$={{x}^{2}}{{e}^{x}}-2x{{e}^{x}}+2{{e}^{x}}+C$

$={{e}^{x}}\left( {{x}^{2}}-2x+2 \right)+C$

4. $x\log x$

Ans:  Let $I=\int{x}\log xdx$

Consider $\text{u}=\log \text{x}\,\,\,\,and\,\,\,\text{v}=\text{x}$ and integrating by parts, to obtain

$I=\log x\int{x}dx-\int{\left\{ \left( \dfrac{d}{dx}\log x \right)\int{x}dx \right\}}dx$

$=\log x\cdot \dfrac{{{x}^{2}}}{2}-\int{\dfrac{1}{x}}\cdot \dfrac{{{x}^{2}}}{2}dx$

$=\dfrac{{{x}^{2}}\log x}{2}\cdot \sqrt{\dfrac{x}{2}}dx=\dfrac{{{x}^{2}}\log x}{2}-\dfrac{{{x}^{2}}}{4}+C$

5. $x\log 2x$

Ans:  Let $I=\int{x}\log 2xdx$

Consider $u=\log 2x$ and $v=x$and integrating by parts, to obtain

$I=\log 2x\int{x}dx-\int{\left\{ \left( \dfrac{d}{dx}2\log x \right)\int{x}dx \right\}}dx$

$=\log 2x\cdot \dfrac{{{x}^{2}}}{2}-\int{\dfrac{2}{2x}}\cdot \dfrac{{{x}^{2}}}{2}dx$

$=\dfrac{{{x}^{2}}\log 2x}{2}-\int{\dfrac{x}{2}}dx$

Integrating using the power rule

$=\dfrac{{{x}^{2}}\log 2x}{2}-\dfrac{{{x}^{2}}}{4}+C$

6. ${{x}^{2}}\log x$

Ans: Let $I=\int{{{x}^{2}}}\log xdx$

Consider $u=\log x$and $v={{x}^{2}}$ and integrating by parts, to obtain

$I=\log x\int{{{x}^{2}}}dx-\int{\left\{ \left( \dfrac{d}{dx}\log x \right)\int{{{x}^{2}}}dx \right\}}dx$

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

Integrating using the power rule

$=\dfrac{{{x}^{3}}\log x}{3}-\int{\dfrac{{{x}^{2}}}{3}}dx=\dfrac{{{x}^{3}}\log x}{3}-\dfrac{{{x}^{3}}}{9}+C$

7. $x{{\sin }^{-1}}x$

Ans: Let $I=\int{x}{{\sin }^{-1}}xdx$

Consider $u={{\sin }^{-1}}x\,\,and\,\,\,v=x$ and integrating by parts, to obtain

$I={{\sin }^{-1}}x\int{x}dx\int{\left\{ \left( \dfrac{d}{dx}{{\sin }^{-1}}x \right)\int{x}dx \right\}}dx$

$={{\sin }^{-1}}x\left( \dfrac{{{x}^{2}}}{2} \right)-\int{\dfrac{1}{\sqrt{1-{{x}^{2}}}}}\cdot \dfrac{{{x}^{2}}}{2}dx$

$=\dfrac{{{x}^{2}}{{\sin }^{-1}}x}{2}+\dfrac{1}{2}\int{\dfrac{-{{x}^{2}}}{\sqrt{1-{{x}^{2}}}}}dx$

Adding and subtracting by 1

$=\dfrac{{{x}^{2}}{{\sin }^{-1}}x}{2}+\dfrac{1}{2}\int{\left\{ \dfrac{1-{{x}^{2}}}{\sqrt{1-{{x}^{2}}}}-\dfrac{1}{\sqrt{1-{{x}^{2}}}} \right\}}dx$

Simplifying,

$=\dfrac{{{x}^{2}}{{\sin }^{-1}}x}{2}+\dfrac{1}{2}\int{\left\{ \sqrt{1-{{x}^{2}}}-\dfrac{1}{\sqrt{1-{{x}^{2}}}} \right\}}dx$

$=\dfrac{{{x}^{2}}{{\sin }^{-1}}x}{2}+\dfrac{1}{2}\left\{ \int{\sqrt{1-{{x}^{2}}}}dx-\int{\dfrac{1}{\sqrt{1-{{x}^{2}}}}}dx \right\}$

$=\dfrac{{{x}^{2}}{{\sin }^{-1}}x}{2}+\dfrac{1}{2}\left\{ \dfrac{x}{2}\sqrt{1-{{x}^{2}}}+\dfrac{1}{2}{{\sin }^{-1}}x-{{\sin }^{-1}}x \right\}+C$

Simplifying,      $=\dfrac{{{x}^{2}}{{\sin }^{-1}}x}{2}+\dfrac{x}{4}\sqrt{1-{{x}^{2}}}+\dfrac{1}{4}{{\sin }^{-1}}x-\dfrac{1}{2}{{\sin }^{-1}}x+C=\dfrac{1}{4}\left( 2{{x}^{2}}-1 \right){{\operatorname{in}}^{-1}}x+\dfrac{x}{4}\sqrt{1-{{x}^{2}}}+C$

8. $x{{\tan }^{-1}}x$

Ans: Let $I=\int{x}{{\tan }^{-1}}xdx$

Consider $\text{u}={{\tan }^{-1}}\text{x}$ and $\text{v}=\text{x}$and integrating by parts, to obtain

$I={{\tan }^{-1}}x\int{x}dx-\int{\left\{ \left( \dfrac{d}{dx}{{\tan }^{-1}}x \right)\int{x}dx \right\}}dx$

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

Adding and subtracting by -1

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

Simplifying,

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

9. $x{{\cos }^{-1}}x$

Ans: Let $I=\int{x}{{\cos }^{-1}}xdx$

Taking $u={{\cos }^{-1}}x$ and $\text{v}=\text{x}$and integrating by parts, to obtain

$I={{\cos }^{-1}}x\int{x}dx-\int{\left\{ \left( \dfrac{d}{dx}{{\cos }^{-1}}x \right)\int{x}dx \right\}}dx$

$={{\cos }^{-1}}x\dfrac{{{x}^{2}}}{2}-\int{\dfrac{-1}{\sqrt{1-{{x}^{2}}}}}\cdot \dfrac{{{x}^{2}}}{2}dx$

Adding and subtracting by -1

$=\dfrac{{{x}^{2}}{{\cos }^{-1}}x}{2}-\dfrac{1}{2}\int{\dfrac{1-{{x}^{2}}-1}{\sqrt{1-{{x}^{2}}}}}dx$

Simplifying,

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

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

$=\dfrac{{{x}^{2}}{{\cos }^{-1}}x}{2}-\dfrac{1}{2}{{I}_{1}}-\dfrac{1}{2}{{\cos }^{-1}}x....\left( 1 \right)$

Where ${{I}_{1}}=\int{\sqrt{1-{{x}^{2}}}}dx$

$\Rightarrow {{I}_{1}}=x\int{\sqrt{1-{{x}^{2}}}}-\int{\dfrac{d}{dx}}\sqrt{1-{{x}^{2}}}\int{x}dx\Rightarrow {{I}_{1}}=x\sqrt{1-{{x}^{2}}}-\int{\dfrac{-{{x}^{2}}}{\sqrt{1-{{x}^{2}}}}}dx$

$\Rightarrow {{I}_{1}}=x\sqrt{1-{{x}^{2}}}-\int{\dfrac{1-{{x}^{2}}-1}{\sqrt{1-{{x}^{2}}}}}dx\Rightarrow {{I}_{1}}=x\sqrt{1-{{x}^{2}}}-\left\{ \int{\sqrt{1-{{x}^{2}}}}dx+\int{\dfrac{-dx}{\sqrt{1-{{x}^{2}}}}} \right\}$

$\Rightarrow {{I}_{1}}=x\sqrt{1-{{x}^{2}}}-\left\{ {{I}_{1}}+{{\cos }^{-1}}x \right\}\Rightarrow 2{{I}_{1}}=x\sqrt{1-{{x}^{2}}}-{{\cos }^{-1}}x$

$\therefore {{I}_{1}}=\dfrac{x}{2}\sqrt{1-{{x}^{2}}}-\dfrac{1}{2}{{\cos }^{-1}}x$

Substituting in (1), we get

$I=\dfrac{x{{\cos }^{-1}}x}{2}-\dfrac{1}{2}\left( \dfrac{x}{2}\sqrt{1-{{x}^{2}}}-\dfrac{1}{2}{{\cos }^{-1}}x \right)-\dfrac{1}{2}{{\cos }^{-1}}x$

Simplifying,

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

10. ${{\left( {{\sin }^{-1}}x \right)}^{2}}$

Ans: Let $I=\int{{{\left( {{\sin }^{-1}}x \right)}^{2}}}\cdot 1dx$

Consider $\text{u}={{\left( {{\sin }^{-1}}\text{x} \right)}^{2}}$ and $\text{v}=1$and integrating by parts, to obtain

$I=\int{\left( {{\sin }^{-1}}x \right)}\cdot \int{1}dx-\int{\left\{ \dfrac{d}{dx}{{\left( {{\sin }^{-1}}x \right)}^{2}}\cdot \int{1}.dx \right\}}dx$

$={{\left( {{\sin }^{-1}}x \right)}^{2}}x-\int{\dfrac{2{{\sin }^{-1}}x}{\sqrt{1-{{x}^{2}}}}}\cdot xdx$

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

$=x{{\left( {{\sin }^{-1}}x \right)}^{2}}+\left[ {{\sin }^{-1}}x\int{\dfrac{-2x}{\sqrt{1-{{x}^{2}}}}}dx-\int{\left\{ \left( \dfrac{d}{dx}{{\sin }^{-1}}x \right)\int{\dfrac{-2x}{\sqrt{1-{{x}^{2}}}}}dx \right\}}dx \right]$

$=x{{\left( {{\sin }^{-1}}x \right)}^{2}}+\left[ {{\sin }^{-1}}x\cdot 2\sqrt{1-{{x}^{2}}}-\int{\dfrac{1}{\sqrt{1-{{x}^{2}}}}}\cdot 2\sqrt{1-{{x}^{2}}}dx \right]$

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

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

11. $\dfrac{x{{\cos }^{-1}}x}{\sqrt{1-{{x}^{2}}}}$

Ans:  Let $I=\int{\dfrac{x{{\cos }^{-1}}x}{\sqrt{1-{{x}^{2}}}}}dx$

Multiplying and dividing by 2

$I=\dfrac{-1}{2}\int{\dfrac{-2x}{\sqrt{1-{{x}^{2}}}}}\cdot {{\cos }^{-1}}xdx$

Consider  $\text{u}={{\cos }^{-1}}\text{x}$ and $\text{v}=\left( \dfrac{-2x}{\sqrt{1-{{x}^{2}}}} \right)$and integrating by parts, to obtain

$I=\dfrac{-1}{2}\left[ {{\cos }^{-1}}x\int{\dfrac{-2x}{\sqrt{1-{{x}^{2}}}}}dx-\int{\left\{ \left( \dfrac{d}{dx}{{\cos }^{-1}}x \right)\int{\dfrac{-2x}{\sqrt{1-{{x}^{2}}}}}dx \right\}}dx \right]$

$=\dfrac{-1}{2}\left[ {{\cos }^{-1}}x\cdot 2\sqrt{1-{{x}^{2}}}-\int{\dfrac{-1}{\sqrt{1-{{x}^{2}}}}}\cdot 2\sqrt{1-{{x}^{2}}}dx \right]=\dfrac{-1}{2}\left[ 2\sqrt{1-{{x}^{2}}}{{\cos }^{-1}}x+\int{2}dx \right]$

Simplifying,

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

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

12. $x{{\sec }^{2}}x$

Ans: Let $I=\int{x}{{\sec }^{2}}xdx$

Consider$\text{u}=\text{x}$ and $\text{v}={{\sec }^{2}}\text{x}$ and integrating by parts, to obtain

$I=x\int{{{\sec }^{2}}}xdx-\int{\left\{ \left\{ \dfrac{d}{dx}x \right\}\int{{{\sec }^{2}}}xdx \right\}}dx$

$=x\tan x-\int{1}\cdot \tan xdx$

$=x\tan x+\log |\cos x|+C$

13. ${{\tan }^{-1}}x$

Ans: Let $I=\int{1}\cdot {{\tan }^{-1}}xdx$

Consider $\text{u}={{\tan }^{-1}}\text{x}$ and $\text{v}=1$ and integrating by parts, to obtain

$I={{\tan }^{-1}}x\int{1}dx-\int{\left\{ \left( \dfrac{d}{dx}{{\tan }^{-1}}x \right)\int{1}.dx \right\}}dx={{\tan }^{-1}}xx-\int{\dfrac{1}{1+{{x}^{2}}}}xd$

$=x{{\tan }^{-1}}x-\dfrac{1}{2}\int{\dfrac{2x}{1+{{x}^{2}}}}dx$

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

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

14. $x{{\left( \log x \right)}^{2}}dx$

Ans:  $I=\int{x}{{(\log x)}^{2}}dx$

Consider$u={{(\log x)}^{2}}$ and $v=1$ and integrating by parts, to obtain

$I={{(\log )}^{2}}\int{x}dx-\int{\left[ \left\{ {{\left( \dfrac{d}{dx}\log x \right)}^{2}} \right\}\int{x}dx \right]}dx$

$=\dfrac{{{x}^{2}}}{2}{{(\log x)}^{2}}-\left[ \int{2}\log x\cdot \dfrac{1}{x}\cdot \dfrac{{{x}^{2}}}{2}dx \right]$

$=\dfrac{{{x}^{2}}}{2}{{(\log x)}^{2}}-\int{x}\log xdx$

Again using integration by parts, to obtain

$I=\dfrac{{{x}^{2}}}{2}{{(\log x)}^{2}}-\left[ \log x\int{x}dx-\int{\left\{ \left( \dfrac{d}{dx}\log x \right)\int{x}dx \right\}}dx \right]$

$=\dfrac{{{x}^{2}}}{2}{{(\log x)}^{2}}-\left[ \dfrac{{{x}^{2}}}{2}-\log x-\int{\dfrac{1}{x}}\cdot \dfrac{{{x}^{2}}}{2}dx \right]$

$=\dfrac{{{x}^{2}}}{2}{{(\log x)}^{2}}-\dfrac{{{x}^{2}}}{2}\log x+\dfrac{1}{2}\int{x}dx=\dfrac{{{x}^{2}}}{2}{{(\log x)}^{2}}-\dfrac{{{x}^{2}}}{2}\log x+\dfrac{{{x}^{2}}}{4}+C$

15. $\left( {{x}^{2}}+1 \right)\log x$

Ans: Let $I=\int{\left( {{x}^{2}}+1 \right)}\log xdx=\int{{{x}^{2}}}\log xdx+\int{\log }xdx$

Let $\text{I}={{\text{I}}_{1}}+{{\text{I}}_{2}}\ldots (1)$

Where,${{I}_{1}}=\int{{{x}^{2}}}\log xdx\,\,\,\,\,\,and\,\,\,{{\text{I}}_{2}}=\int{\log }xdx$

${{I}_{1}}=\int{{{x}^{2}}}\log xdx$ 

Consider$\text{u}=\log \text{x}$ and $v=x^2$ and integrating by parts, to obtain

${{I}_{1}}=\log x-\int{{{x}^{2}}}dx-\int{\left\{ \left( \dfrac{d}{dx}\log x \right)\int{{{x}^{2}}}dx \right\}}dx$

$=\log x\cdot \dfrac{{{x}^{3}}}{3}-\int{\dfrac{1}{x}}\cdot \dfrac{{{x}^{3}}}{3}dx=\dfrac{{{x}^{3}}}{3}\log x-\dfrac{1}{3}\int{{{x}^{2}}}dx$

$=\dfrac{{{x}^{3}}}{3}\log x-\dfrac{{{x}^{3}}}{9}+{{C}_{1}}\quad \ldots (2)$

${{I}_{2}}=\int{\log }xdx$

Consider $\text{u}=\log \text{x}$ and $\text{v}=1$and integrating by parts, to obtain

${{I}_{2}}=\log x\int{1}.dx-\int{\left\{ \left( \dfrac{d}{dx}\log x \right)\int{1}.dx \right\}}$

$=\log x\cdot x-\int{\dfrac{1}{x}}xdx$

$=x\log x-x..\left( 3 \right)$

Using equations (2) and (3) in (1), we get

$I=\dfrac{{{x}^{3}}}{3}\log x-\dfrac{{{x}^{3}}}{9}+{{C}_{1}}+x\log x-x+{{C}_{2}}$

$=\dfrac{{{x}^{3}}}{3}\log x-\dfrac{{{x}^{3}}}{9}+x\log x-x+\left( {{C}_{1}}+{{C}_{2}} \right)$

$=\left( \dfrac{{{x}^{3}}}{3}+x \right)\log x-\dfrac{{{x}^{3}}}{9}-x+C$

16. ${{e}^{x}}\left( \sin x+\cos x \right)$

Ans: Consider$I=\int{{{e}^{x}}}(\sin x+\cos x)dx$

Consider$f(x)=\sin x$

${{f}^{\prime }}(x)=\cos x$

$I=\int{{{e}^{x}}}\left\{ f(x)+{{f}^{\prime }}(x) \right\}dx$

Since, $\int{{{e}^{x}}}\left\{ f(x)+{{f}^{\prime }}(x) \right\}dx={{e}^{x}}f(x)+C$

$\therefore I={{e}^{x}}\sin x+C$

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

Ans:  Consider $I=\int{\dfrac{x{{e}^{x}}}{{{(1+x)}^{2}}}}dx=\int{{{e}^{x}}}\left\{ \dfrac{x}{{{(1+x)}^{2}}} \right\}dx$

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

Here, $f(x)=\dfrac{1}{1+x}\quad {{f}^{\prime }}(x)=\dfrac{-1}{{{(1+x)}^{2}}}$

$\Rightarrow \int{\dfrac{x{{e}^{x}}}{{{(1+x)}^{2}}}}dx=\int{{{e}^{x}}}\left\{ f(x)+{{f}^{\prime }}(x) \right\}dx$

Since, $\int{{{e}^{x}}}\left\{ f(x)+{{f}^{\prime }}(x) \right\}dx={{e}^{x}}f(x)+C$

$\therefore \int{\dfrac{x{{e}^{x}}}{{{(1+x)}^{2}}}}dx=\dfrac{{{e}^{x}}}{1+x}+C$

18. Integrate the function - ${{e}^{x}}\left( \dfrac{1+\sin x}{1+\cos x} \right)$

Ans: First simplify –${{e}^{x}}\left( \dfrac{1+\sin x}{1+\cos x} \right)$

It is known that – 

$1+\sin x={{\sin }^{2}}\dfrac{x}{2}+{{\cos }^{2}}\dfrac{x}{2}+2\sin \dfrac{x}{2}\cos \dfrac{x}{2}$

$1+\cos x=2{{\cos }^{2}}\dfrac{x}{2}$

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

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

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

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

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

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

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

But, $1+{{\tan }^{2}}\dfrac{x}{2}={{\sec }^{2}}\dfrac{x}{2}$

$=\dfrac{1}{2}{{e}^{x}}\left( {{\sec }^{2}}\dfrac{x}{2}+2\tan \dfrac{x}{2} \right)$

$={{e}^{x}}\left( \dfrac{1}{2}{{\sec }^{2}}\dfrac{x}{2}+\tan \dfrac{x}{2} \right)$

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

It is known that, $\int{{{e}^{x}}\left\{ f(x)+f'(x) \right\}dx=}{{e}^{x}}f(x)+C$

If we say, $f(x)=\tan \dfrac{x}{2}\Rightarrow f'(x)=\dfrac{1}{2}{{\sec }^{2}}\dfrac{x}{2}$

Thus, we get – $\int{{{e}^{x}}\left( \dfrac{1+\sin x}{1+\cos x} \right)}dx={{e}^{x}}\tan \dfrac{x}{2}+C$

19. Integrate the function - ${{e}^{x}}\left( \dfrac{1}{x}-\dfrac{1}{{{x}^{2}}} \right)$

Ans: Say, $I=\int{{{e}^{x}}\left( \dfrac{1}{x}-\dfrac{1}{{{x}^{2}}} \right)dx}$

Suppose, $f(x)=\dfrac{1}{x}\Rightarrow f'(x)=-\dfrac{1}{{{x}^{2}}}$

It is known that, $\int{{{e}^{x}}\left\{ f(x)+f'(x) \right\}dx=}{{e}^{x}}f(x)+C$

Thus, we get – $I=\int{{{e}^{x}}\left( \dfrac{1}{x}-\dfrac{1}{{{x}^{2}}} \right)dx}=\dfrac{{{e}^{x}}}{x}+C$

20. Integrate the function - $\dfrac{(x-3){{e}^{x}}}{{{(x-1)}^{3}}}$

Ans: $\int{{{e}^{x}}\dfrac{(x-3)}{{{(x-1)}^{3}}}dx=\int{{{e}^{x}}\left[ \dfrac{(x-1-2)}{{{(x-1)}^{3}}} \right]dx}}$

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

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

Suppose, $f(x)=\dfrac{1}{{{(x-1)}^{2}}}\Rightarrow f'(x)=-\dfrac{2}{{{(x-1)}^{3}}}$

It is known that, $\int{{{e}^{x}}\left\{ f(x)+f'(x) \right\}dx=}{{e}^{x}}f(x)+C$

Thus, $\int{{{e}^{x}}\dfrac{(x-3)}{{{(x-1)}^{3}}}dx=\dfrac{{{e}^{x}}}{{{(x-1)}^{2}}}+C}$

21. Integrate the function - ${{e}^{2x}}\sin x$

Ans: Say,  $I=\int{{{e}^{2x}}\sin xdx}$

Perform Integration by parts – $\int{uv}dx=u\int{vdx}-\int{\left( u'\int{vdx} \right)dx}$

With –$u=\sin x\text{   }v={{e}^{2x}}$

$I=\int{{{e}^{2x}}\sin x}dx=\sin x\int{{{e}^{2x}}dx}-\int{\left[ \left( \dfrac{d}{dx}\sin x \right)\int{{{e}^{2x}}dx} \right]dx}$

$=\sin x\dfrac{{{e}^{2x}}}{2}-\int{\left[ \left( \cos x \right)\dfrac{{{e}^{2x}}}{2} \right]dx}$

$=\sin x\dfrac{{{e}^{2x}}}{2}-\dfrac{1}{2}\int{\left( {{e}^{2x}}\cos x \right)dx}$

Perform Integration by parts for – $\int{\left( {{e}^{2x}}\cos x \right)dx}$

$=\sin x\dfrac{{{e}^{2x}}}{2}-\dfrac{1}{2}\left\{ \cos x\int{{{e}^{2x}}dx}-\int{\left[ \left( \dfrac{d}{dx}\cos x \right)\int{{{e}^{2x}}dx} \right]dx} \right\}$

$=\sin x\dfrac{{{e}^{2x}}}{2}-\dfrac{1}{2}\left\{ \cos x\dfrac{{{e}^{2x}}}{2}-\int{\left[ \left( -\sin x \right)\dfrac{{{e}^{2x}}}{2} \right]dx} \right\}$

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

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

But, $I=\int{{{e}^{2x}}\sin xdx}$

$\Rightarrow I=\sin x\dfrac{{{e}^{2x}}}{2}-\dfrac{{{e}^{2x}}\cos x}{4}-\dfrac{1}{4}I$

$\Rightarrow I+\dfrac{1}{4}I=\sin x\dfrac{{{e}^{2x}}}{2}-\dfrac{{{e}^{2x}}\cos x}{4}$

$\Rightarrow \dfrac{5}{4}I=\dfrac{{{e}^{2x}}\sin x}{2}-\dfrac{{{e}^{2x}}\cos x}{4}$

$\Rightarrow \dfrac{5}{4}I=\dfrac{2{{e}^{2x}}\sin x}{4}-\dfrac{{{e}^{2x}}\cos x}{4}$

$\Rightarrow 5I={{e}^{2x}}(2\sin x-\cos x)$

Thus, we get – $I=\dfrac{{{e}^{2x}}}{5}(2\sin x-\cos x)+C$.

22. Integrate the function - ${{\sin }^{-1}}\left( \dfrac{2x}{1+{{x}^{3}}} \right)$

Ans: Say, $x=\tan \theta \text{ }\Rightarrow \text{dx=se}{{\text{c}}^{2}}\theta d\theta $

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

But, $\sin 2\theta =\dfrac{2\tan \theta }{1+{{\tan }^{3}}\theta }$

$\Rightarrow {{\sin }^{-1}}\left( \dfrac{2x}{1+{{x}^{3}}} \right)={{\sin }^{-1}}\left( \dfrac{2\tan \theta }{1+{{\tan }^{3}}\theta } \right)=\therefore {{\sin }^{-1}}\left( \dfrac{2x}{1+{{x}^{3}}} \right)={{\sin }^{-1}}\left( \sin 2\theta  \right)=2\theta $

Therefore, $\int{{{\sin }^{-1}}\left( \dfrac{2x}{1+{{x}^{3}}} \right)dx}=\int{2\theta \text{se}{{\text{c}}^{2}}\theta d\theta }$

$=2\int{\theta \text{se}{{\text{c}}^{2}}\theta d\theta }$

Perform Integration by parts – $\int{uv}dx=u\int{vdx}-\int{\left( u'\int{vdx} \right)dx}$

With –$u=\theta \text{   }v={{\sec }^{2}}\theta $

$2\int{\theta \text{se}{{\text{c}}^{2}}\theta d\theta }=2\left\{ \theta \int{{{\sec }^{2}}\theta d\theta }-\int{\left[ \left( \dfrac{d}{d\theta }\theta  \right)\int{{{\sec }^{2}}\theta d\theta } \right]d\theta } \right\}$

$=2\left\{ \theta \tan \theta -\int{\left[ \tan \theta  \right]d\theta } \right\}$

$=2\left\{ \theta \tan \theta -(-\log |\cos \theta |) \right\}+C$

$=2\left\{ \theta \tan \theta +\log |\cos \theta | \right\}+C$

Replace $\theta ={{\tan }^{-1}}x$

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

It is known that – ${{\tan }^{-1}}x={{\cos }^{-1}}\dfrac{1}{\sqrt{1+{{x}^{2}}}}$

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

$=2\left\{ x{{\tan }^{-1}}x+\log |\dfrac{1}{\sqrt{1+{{x}^{2}}}}| \right\}+C$

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

Here, $\log {{m}^{n}}=n\log m$

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

$=2x{{\tan }^{-1}}x-\log (1+{{x}^{2}})+C$

Thus, $\int{{{\sin }^{-1}}\left( \dfrac{2x}{1+{{x}^{3}}} \right)dx}=2x{{\tan }^{-1}}x-\log (1+{{x}^{2}})+C$

23. Choose the correct answer: $\int{{{x}^{2}}{{e}^{{{x}^{3}}}}}dx$ equals

1. $\dfrac{1}{3}{{e}^{{{x}^{3}}}}+C$

2. $\dfrac{1}{3}{{e}^{{{x}^{2}}}}+C$

3. $\dfrac{1}{2}{{e}^{{{x}^{3}}}}+C$

4. $\dfrac{1}{2}{{e}^{{{x}^{2}}}}+C$

Ans: Say, $I=\int{{{x}^{2}}{{e}^{{{x}^{3}}}}}dx$

Suppose, $t={{x}^{3}}\Rightarrow dt=3{{x}^{2}}dx$

Rewriting the equation – $I=\int{{{x}^{2}}{{e}^{{{x}^{3}}}}}dx=\dfrac{1}{3}\int{{{e}^{t}}}dt$

$\Rightarrow I=\dfrac{1}{3}\int{{{e}^{t}}}dt=\dfrac{1}{3}{{e}^{t}}+C$

Replacing $t={{x}^{3}}$

$\Rightarrow I=\dfrac{1}{3}{{e}^{{{x}^{3}}}}+C$

The correct option is A.

24. Choose the correct answer: $\int{{{e}^{x}}\sec x(1+\tan x)}dx$

1. ${{e}^{x}}\cos x+C$

2. ${{e}^{x}}\sec x+C$

3. ${{e}^{x}}\sin x+C$

4. ${{e}^{x}}\tan x+C$

Ans: Say, $I=\int{{{e}^{x}}\sec x(1+\tan x)}dx$

$\Rightarrow I=\int{{{e}^{x}}(\sec x+\sec x\tan x)}dx$

Suppose, $f(x)=\sec x\Rightarrow f'(x)=\sec x\tan x$

It is known that, $\int{{{e}^{x}}\left\{ f(x)+f'(x) \right\}dx=}{{e}^{x}}f(x)+C$

$\Rightarrow I=\int{{{e}^{x}}(\sec x+\sec x\tan x)}dx={{e}^{x}}\sec x+C$

Thus, $I={{e}^{x}}\sec x+C$

The correct option is B.

Conclusion

NCERT for Chapter 7 Exercise 7.6 in Class 12 Mathematics, as provided by Vedantu, has been a crucial learning experience. Through solving a series of problems in class 12 maths ex 7.6 involving integration of rational functions, we've grasped fundamental techniques like partial fraction decomposition and trigonometric substitutions. Understanding these methods is key to mastering integration. It's essential to focus on practicing these methods extensively to build confidence and proficiency. By doing so, we not only enhance our mathematical skills but also lay a solid foundation for tackling more complex problems in the future.

Class 12 Maths Chapter 7: Exercises Breakdown

CBSE Class 12 Maths Chapter 7 Other Study Materials

NCERT Solutions for Class 12 Maths | Chapter-wise List

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