NCERT Solutions Class 12 maths Chapter-7 (Integrals)Exercise 7.6

NCERT Solutions Class 12 Maths from class 12th Students will get the answers of Chapter-7 (Integrals)Exercise 7.6 This chapter will help you to learn the basics and you should expect at least one question in your exam from this chapter.
We have given the answers of all the questions of NCERT Board Mathematics Textbook in very easy language, which will be very easy for the students to understand and remember so that you can pass with good marks in your examination.

Exercise 7.6

Q1. Integrate the function :$x\sin x$

Answer. Let $I=\int x\sin xdx$ Taking x as first function and sin x as second function and integrating by parts, we obtain $I=x\int \sin xdx-\int \{\left(\frac{d}{dx}x\right)\int \sin xdx\}dx$ $\begin{array}{c}=x(-\cos x)-\int 1\cdot (-\cos x)dx\\ =-x\cos x+\sin x+C\end{array}$

Q2. Integrate the function :$x\sin 3x$

Answer. Let $I=\int x\sin 3xdx$ Taking x as first function and sin 3x as second function and integrating by parts, we obtain $I=x\int \sin 3xdx-\int \{\left(\frac{d}{dx}x\right)\int \sin 3xdx\}$ $\begin{array}{cc}& =x\left(\frac{-\cos 3x}{3}\right)-\int 1\cdot \left(\frac{-\cos 3x}{3}\right)dx\\ & =\frac{-x\cos 3x}{3}+\frac{1}{3}\int \cos 3xdx\\ & =\frac{-x\cos 3x}{3}+\frac{1}{9}\sin 3x+C\end{array}$

Q3. Integrate the function :$x^2{e}^x$

Answer. Let $I=\int x^2{e}^{x}dx$ Taking $x^2$ as first function and $e^x$ as second function and integrating by parts, we obtain $\begin{array}{cc}I& =x^2\int e^{x}dx-\int \{\left(\frac{d}{dx}{x}^2\right)\int e^{x}dx\}dx\\ & =x^2{e}^x-\int 2x\cdot e^{x}dx\\ & =x^2{e}^x-2\int x\cdot e^{x}dx\end{array}$ Again integrating by parts, we obtain $\begin{array}{cc}& =x^2{e}^x-2[x\cdot \int e^{x}dx-\int \{\left(\frac{d}{dx}x\right)\cdot \int e^{x}dx\}dx]\\ & =x^2{e}^x-2[x{e}^x-\int e^{x}dx]\\ & =x^2{e}^x-2[x{e}^x-e^x]\\ & =x^2{e}^x-2x{e}^x+2e^x+C\\ & =e^x(x^2-2x+2)+C\end{array}$

Q4. Integrate the function :$x\log x$

Answer. Let $I=\int x\log xdx$ Taking log x as first function and x as second function and integrating by parts, we obtain $\begin{array}{cc}I& =\log x\int xdx-\int \{\left(\frac{d}{dx}\log x\right)\int xdx\}dx\\ & =\log x\cdot \frac{x^2}{2}-\int \frac{1}{x}\cdot \frac{x^2}{2}dx\\ & =\frac{x^2\log x}{2}-\int \frac{x}{2}dx\\ & =\frac{x^2\log x}{2}-\frac{x^2}{4}+C\end{array}$

Q5. Integrate the function :$x\log 2x$

Answer. Let $I=\int x\log 2xdx$ Taking log 2x as first function and x as second function and integrating by parts, we obtain $\begin{array}{cc}I& =\log 2x\int xdx-\int \{\left(\frac{d}{dx}2\log x\right)\int xdx\}dx\\ & =\log 2x\cdot \frac{x^2}{2}-\int \frac{2}{2x}\cdot \frac{x^2}{2}dx\\ & =\frac{x^2\log 2x}{2}-\int \frac{x}{2}dx\\ & =\frac{x^2\log 2x}{2}-\frac{x^2}{4}+C\end{array}$

Q6. Integrate the function :$x^2\log x$

Answer. Let $I=\int x^2\log xdx$ Taking log x as first function and $x^2$ as second function and integrating by parts, we obtain $\begin{array}{cc}I& =\log x\int x^{2}dx-\int \{\left(\frac{d}{dx}\log x\right)\int x^{2}dx\}dx\\ & =\log x\left(\frac{x^3}{3}\right)-\int \frac{1}{x}\cdot \frac{x^3}{3}dx\\ & =\frac{x^3\log x}{3}-\int \frac{x^2}{3}dx\\ & =\frac{x^3\log x}{3}-\frac{x^3}{9}+C\end{array}$

Q7. Integrate the function :$x{\sin }^{-1}x$

Answer. Let $I=\int x{\sin }^{-1}xdx$ Taking ${\sin }^{-1}x$ as first function and x as second function and integrating by parts, we obtain $I={\sin }^{-1}x\int xdx-\int \{\left(\frac{d}{dx}{\sin }^{-1}x\right)\int xdx\}dx$ $={\sin }^{-1}x\left(\frac{x^2}{2}\right)-\int \frac{1}{\sqrt{1-x^2}}\cdot \frac{x^2}{2}dx$ $\begin{array}{c}=\frac{x^2{\sin }^{-1}x}{2}+\frac{1}{2}\int \frac{-x^2}{\sqrt{1-x^2}}dx\\ =\frac{x^2{\sin }^{-1}x}{2}+\frac{1}{2}\int \{\frac{1-x^2}{\sqrt{1-x^2}}-\frac{1}{\sqrt{1-x^2}}\}dx\end{array}$ $\begin{array}{c}=\frac{x^2{\sin }^{-1}x}{2}+\frac{1}{2}\int \{\sqrt{1-x^2}-\frac{1}{\sqrt{1-x^2}}\}dx\\ =\frac{x^2{\sin }^{-1}x}{2}+\frac{1}{2}\{\int \sqrt{1-x^2}dx-\int \frac{1}{\sqrt{1-x^2}}dx\}\\ =\frac{x^2{\sin }^{-1}x}{2}+\frac{1}{2}\{\frac{x}{2}\sqrt{1-x^2}+\frac{1}{2}{\sin }^{-1}x-{\sin }^{-1}x\}+C\\ =\frac{x^2{\sin }^{-1}x}{2}+\frac{x}{4}\sqrt{1-x^2}+\frac{1}{4}{\sin }^{-1}x-\frac{1}{2}{\sin }^{-1}x+C\end{array}$ $=\frac{1}{4}(2x^2-1){\sin }^{-1}x+\frac{x}{4}\sqrt{1-x^2}+C$

Q8. Integrate the function :$x{\tan }^{-1}x$

Answer. Let $I=\int x{\tan }^{-1}xdx$ Taking ${\tan }^{-1}x$ as first function and x as second function and integrating by parts, we obtain $\begin{array}{cc}I& ={\tan }^{-1}x\int xdx-\int \{\left(\frac{d}{dx}{\tan }^{-1}x\right)\int xdx\}dx\\ & ={\tan }^{-1}x\left(\frac{x^2}{2}\right)-\int \frac{1}{1+x^2}\cdot \frac{x^2}{2}dx\\ & =\frac{x^2{\tan }^{-1}x}{2}-\frac{1}{2}\int \frac{x^2}{1+x^2}dx\end{array}$ $\begin{array}{cc}& =\frac{x^2{\tan }^{-1}x}{2}-\frac{1}{2}\int (\frac{x^2+1}{1+x^2}-\frac{1}{1+x^2})dx\\ & =\frac{x^2{\tan }^{-1}x}{2}-\frac{1}{2}\int (1-\frac{1}{1+x^2})dx\\ & =\frac{x^2{\tan }^{-1}x}{2}-\frac{1}{2}(x-{\tan }^{-1}x)+C\\ & =\frac{x^2}{2}{\tan }^{-1}x-\frac{x}{2}+\frac{1}{2}{\tan }^{-1}x+C\end{array}$

Q9. Integrate the function :$x{\cos }^{-1}x$

Answer. Let $I=\int x{\cos }^{-1}xdx$ Taking  as first function and x as second function and integrating by parts, we obtain $\begin{array}{cc}I& ={\cos }^{-1}x\int xdx-\int \{\left(\frac{d}{dx}{\cos }^{-1}x\right)\int xdx\}dx\\ & ={\cos }^{-1}x\frac{x^2}{2}-\int \frac{-1}{\sqrt{1-x^2}}\cdot \frac{x^2}{2}dx\\ & =\frac{x^2{\cos }^{-1}x}{2}-\frac{1}{2}\int \frac{1-x^2-1}{\sqrt{1-x^2}}dx\\ & =\frac{x^2{\cos }^{-1}x}{2}-\frac{1}{2}\int \{\sqrt{1-x^2}+\left(\frac{-1}{\sqrt{1-x^2}}\right)\}dx\\ & =\frac{x^2{\cos }^{-1}x}{2}-\frac{1}{2}\int \sqrt{1-x^2}dx-\frac{1}{2}\int \left(\frac{-1}{\sqrt{1-x^2}}\right)dx\end{array}$ $\begin{array}{c}=\frac{x^2{\cos }^{-1}x}{2}-\frac{1}{2}{I}_1-\frac{1}{2}{\cos }^{-1}x\\ \text{where,}{I}_1=\int \sqrt{1-x^2}dx\\ \Rightarrow I_1=x\sqrt{1-x^2}-\int \frac{d}{dx}\sqrt{1-x^2}\int xdx\\ \Rightarrow I_1=x\sqrt{1-x^2}-\int \frac{-2x}{2\sqrt{1-x^2}}\cdot xdx\end{array}$ $\begin{array}{c}=\frac{x^2{\cos }^{-1}x}{2}-\frac{1}{2}\int \frac{1-x^2-1}{\sqrt{1-x^2}}dx\\ =\frac{x^2{\cos }^{-1}x}{2}-\frac{1}{2}\int \{\sqrt{1-x^2}+\left(\frac{-1}{\sqrt{1-x^2}}\right)\}dx\\ =\frac{x^2{\cos }^{-1}x}{2}-\frac{1}{2}\int \sqrt{1-x^2}dx-\frac{1}{2}\int \left(\frac{-1}{\sqrt{1-x^2}}\right)dx\\ =\frac{x^2{\cos }^{-1}x}{2}-\frac{1}{2}{I}_1-\frac{1}{2}{\cos }^{-1}x\end{array}$ $\begin{array}{cc}& =\frac{x^2{\cos }^{-1}x}{2}-\frac{1}{2}\int \{\sqrt{1-x^2}+\left(\frac{-1}{\sqrt{1-x^2}}\right)\}dx\\ & =\frac{x^2{\cos }^{-1}x}{2}-\frac{1}{2}\int \sqrt{1-x^2}dx-\frac{1}{2}\int \left(\frac{-1}{\sqrt{1-x^2}}\right)dx\\ & =\frac{x^2{\cos }^{-1}x}{2}-\frac{1}{2}{I}_1-\frac{1}{2}{\cos }^{-1}x\end{array}$ $\begin{array}{c}\text{where,}{I}_1=\int \sqrt{1-x^2}dx\\ \Rightarrow I_1=x\sqrt{1-x^2}-\int \frac{d}{dx}\sqrt{1-x^2}\int xdx\\ \Rightarrow I_1=x\sqrt{1-x^2}-\int \frac{-2x}{2\sqrt{1-x^2}}\cdot xdx\\ \Rightarrow I_1=x\sqrt{1-x^2}-\int \frac{-x^2}{\sqrt{1-x^2}}dx\\ \Rightarrow I_1=x\sqrt{1-x^2}-\int \frac{1-x^2-1}{\sqrt{1-x^2}}dx\end{array}$ $\begin{array}{c}\Rightarrow I_1=x\sqrt{1-x^2}-\int \frac{1-x-1}{\sqrt{1-x^2}}dx\\ \Rightarrow I_1=x\sqrt{1-x^2}-\{\int \sqrt{1-x^2}dx+\int \frac{-dx}{\sqrt{1-x^2}}\}\\ \Rightarrow I_1=x\sqrt{1-x^2}-\{I_1+{\cos }^{-1}x\}\\ \Rightarrow 2I_1=x\sqrt{1-x^2}-{\cos }^{-1}x\\ \therefore I_1=\frac{x}{2}\sqrt{1-x^2}-\frac{1}{2}{\cos }^{-1}x\end{array}$ Substituting in (1) , we obtain $\begin{array}{cc}I& =\frac{x{\cos }^{-1}x}{2}-\frac{1}{2}(\frac{x}{2}\sqrt{1-x^2}-\frac{1}{2}{\cos }^{-1}x)-\frac{1}{2}{\cos }^{-1}x\\ & =\frac{(2x^2-1)}{4}{\cos }^{-1}x-\frac{x}{4}\sqrt{1-x^2}+C\end{array}$

Q10. Integrate the function :${\left({\sin }^{-1}x\right)}^2$

Answer. Let $I=\int {\left({\sin }^{-1}x\right)}^2\cdot 1dx$ Taking ${\left({\sin }^{-1}x\right)}^2$ as first function and 1 as second function and integrating by parts, we obtain $\begin{array}{cc}I& =\left({\sin }^{-1}x\right)\int 1dx-\int \{\frac{d}{dx}{\left({\sin }^{-1}x\right)}^2\cdot \int 1\cdot dx\}dx\\ & ={\left({\sin }^{-1}x\right)}^2\cdot x-\int \frac{2{\sin }^{-1}x}{\sqrt{1-x^2}}\cdot xdx\\ & =x{\left({\sin }^{-1}x\right)}^2+\int {\sin }^{-1}x\cdot \left(\frac{-2x}{\sqrt{1-x^2}}\right)dx\end{array}$ $\begin{array}{c}=x{\left({\sin }^{-1}x\right)}^2+[{\sin }^{-1}x\int \frac{-2x}{\sqrt{1-x^2}}dx-\int \{\left(\frac{d}{dx}{\sin }^{-1}x\right)\int \frac{-2x}{\sqrt{1-x^2}}dx\}dx]\\ =x{\left({\sin }^{-1}x\right)}^2+[{\sin }^{-1}x\cdot 2\sqrt{1-x^2}-\int \frac{1}{\sqrt{1-x^2}}\cdot 2\sqrt{1-x^2}dx]\end{array}$ $\begin{array}{c}=x{\left({\sin }^{-1}x\right)}^2+2\sqrt{1-x^2}{\sin }^{-1}x-\int 2dx\\ =x{\left({\sin }^{-1}x\right)}^2+2\sqrt{1-x^2}{\sin }^{-1}x-2x+C\end{array}$

Q11. Integrate the function :$\frac{x{\cos }^{-1}x}{\sqrt{1-x^2}}$

Answer. Let $I=\int \frac{x{\cos }^{-1}x}{\sqrt{1-x^2}}dx$ $I=\frac{-1}{2}\int \frac{-2x}{\sqrt{1-x^2}}\cdot {\cos }^{-1}xdx$ Taking ${\cos }^{-1}x$ as first function and $\left(\frac{-2x}{\sqrt{1-x^2}}\right)$ as second function and integrating by parts, we obtain $\begin{array}{cc}I& =\frac{-1}{2}[{\cos }^{-1}x\int \frac{-2x}{\sqrt{1-x^2}}dx-\int \{\left(\frac{d}{dx}{\cos }^{-1}x\right)\int \frac{-2x}{\sqrt{1-x^2}}dx\}dx]\\ & =\frac{-1}{2}[{\cos }^{-1}x\cdot 2\sqrt{1-x^2}-\int \frac{-1}{\sqrt{1-x^2}}\cdot 2\sqrt{1-x^2}dx]\\ & =\frac{-1}{2}[2\sqrt{1-x^2}{\cos }^{-1}x+\int 2dx]\\ & =\frac{-1}{2}[2\sqrt{1-x^2}{\cos }^{-1}x+2x]+C\\ & =-[\sqrt{1-x^2}{\cos }^{-1}x+x]+C\end{array}$

Q12. Integrate the function :$x{\sec }^{2}x$

Answer. Let $I=\int x{\sec }^{2}xdx$ Taking x as first function and ${\sec }^{2}x$ as second function and integrating by parts, we obtain $\begin{array}{cc}I& =x\int {\sec }^{2}xdx-\int \{\left\{\frac{d}{dx}x\right\}\int {\sec }^{2}xdx\}dx\\ & =x\tan x-\int 1\cdot \tan xdx\\ & =x\tan x+\log |\cos x|+C\end{array}$

Q13. Integrate the function :${\tan }^{-1}x$

Answer. Let $I=\int 1\cdot {\tan }^{-1}xdx$ Taking ${\tan }^{-1}x$ as first function and 1 as second function and integrating by parts, we obtain $I={\tan }^{-1}x\int 1dx-\int \{\left(\frac{d}{dx}{\tan }^{-1}x\right)\int 1\cdot dx\}dx$ $\begin{array}{c}={\tan }^{-1}x\cdot x-\int \frac{1}{1+x^2}\cdot xdx\\ =x{\tan }^{-1}x-\frac{1}{2}\int \frac{2x}{1+x^2}dx\end{array}$ $\begin{array}{c}=x{\tan }^{-1}x-\frac{1}{2}\log |1+x^2|+C\\ =x{\tan }^{-1}x-\frac{1}{2}\log (1+x^2)+C\end{array}$

Q14. Integrate the function :$x(\log x{)}^2$

Answer. $I=\int x(\log x{)}^{2}dx$ Taking $(\log x{)}^2$ as first function and 1 as second function and integrating by parts, we obtain $\begin{array}{cc}I& =(\log x{)}^2\int xdx-\int \left\{{\left(\frac{d}{dx}\log x\right)}^{-}\right\}\int xdx]dx\\ & =\frac{x^2}{2}(\log x{)}^2-[2\log x\cdot \frac{1}{x}\cdot \frac{x^2}{2}dx]\\ & =\frac{x^2}{2}(\log x{)}^2-\int x\log xdx\end{array}$ Again integrating by parts , we obtain $\begin{array}{cc}I& =\frac{x^2}{2}(\log x{)}^2-[\log x\int xdx-\int \{\left(\frac{d}{dx}\log x\right)\int xdx\}dx]\\ & =\frac{x^2}{2}(\log x{)}^2-[\frac{x^2}{2}-\log x-\int \frac{1}{x}\cdot \frac{x^2}{2}dx]\\ & =\frac{x^2}{2}(\log x{)}^2-\frac{x^2}{2}\log x+\frac{1}{2}\int xdx\\ & =\frac{x^2}{2}(\log x{)}^2-\frac{x^2}{2}\log x+\frac{x^2}{4}+C\end{array}$

Q15. Integrate the function :$(x^2+1)\log x$

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Let $I=\int (x^2+1)\log xdx=\int x^2\log xdx+\int \log xdx$ Let $I=I_1+I_2\dots \left(1\right)$ Where $I_1=\int x^2\log xd{x}_{\text{and}}{I}_2=\int \log xdx$ $I_1=\int x^2\log xdx$ Taking log x as first function and $x^2$ as second function and integrating by parts , we obtain $\begin{array}{cc}{I}_1& =\log x-\int x^{2}dx-\int \{\left(\frac{d}{dx}\log x\right)\int x^{2}dx\}dx\\ & =\log x\cdot \frac{x^3}{3}-\int \frac{1}{x}\cdot \frac{x^3}{3}dx\\ & =\frac{x^3}{3}\log x-\frac{1}{3}(\int x^{2}dx)\\ & =\frac{x^3}{3}\log x-\frac{x^3}{9}+C_1\\ I_2& =\int \log xdx\end{array}$ Taking log x as first function and 1 as second function and integrating by parts , we obtain $\begin{array}{cc}{I}_2& =\log x\int 1\cdot dx-\int \{\left(\frac{d}{dx}\log x\right)\int 1\cdot dx\}\\ & =\log x\cdot x-\int \frac{1}{x}\cdot xdx\\ & =x\log x-\int 1dx\\ & =x\log x-x+C_2\end{array}$ Using equations (2) and (3) in (1) , we obtain $\begin{array}{cc}I& =\frac{x^3}{3}\log x-\frac{x^3}{9}+C_1+x\log x-x+C_2\\ & =\frac{x^3}{3}\log x-\frac{x^3}{9}+x\log x-x+(C_1+C_2)\\ & =(\frac{x^3}{3}+x)\log x-\frac{x^3}{9}-x+C\end{array}$

Q16. Integrate the function :$e^x(\sin x+\cos x)$

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 Let $I=\int e^x(\sin x+\cos x)dx$ Let $f\left(x\right)=\sin x$ $\begin{array}{c}{f}^{\prime }\left(x\right)=\cos x\\ I=\int e^x\left\{f\right(x)+f^{\prime }(x\left)\right\}dx\end{array}$ It is known that , $\int e^x\left\{f\right(x)+f^{\prime }(x\left)\right\}dx=e^{x}f\left(x\right)+C$ $\therefore I=e^x\sin x+C$

Q17. Integrate the function :$\frac{x{e}^x}{(1+x{)}^2}$

Answer. Let $I=\int \frac{x{e}^x}{(1+x{)}^2}dx=\int e^x\left\{\frac{x}{(1+x{)}^2}\right\}dx$ $\begin{array}{c}=\int e^x\left\{\frac{1+x-1}{(1+x{)}^2}\right\}dx\\ =\int e^x\{\frac{1}{1+x}-\frac{1}{(1+x{)}^2}\}dx\end{array}$ Let $f\left(x\right)=\frac{1}{1+x}{f}^{\prime }\left(x\right)=\frac{-1}{(1+x{)}^2}$ $\Rightarrow \int \frac{x{e}^x}{(1+x{)}^2}dx=\int e^x\left\{f\right(x)+f^{\prime }(x\left)\right\}dx$ It is known that, $\int e^x\left\{f\right(x)+f^{\prime }(x\left)\right\}dx=e^{x}f\left(x\right)+C$ $\therefore \int \frac{x{e}^x}{(1+x{)}^2}dx=\frac{e^x}{1+x}+C$

Q18. Integrate the function :$e^x\left(\frac{1+\sin x}{1+\cos x}\right)$

Answer. $\begin{array}{c}{e}^x\left(\frac{1+\sin x}{1+\cos x}\right)\\ =e^x\left(\frac{{\sin }^2\frac{x}{2}+{\cos }^2\frac{x}{2}+2\sin \frac{x}{2}\cos \frac{x}{2})}{2{\cos }^2\frac{x}{2}}\right)\\ =\frac{e^x{(\sin \frac{x}{2}+\cos \frac{x}{2})}^2}{2{\cos }^2\frac{x}{2}}\end{array}$ $\begin{array}{cc}& =\frac{1}{2}{e}^x\cdot {\left(\frac{\sin \frac{x}{2}+\cos \frac{x}{2}}{\cos \frac{x}{2}}\right)}^2\\ & =\frac{1}{2}{e}^x{[\tan \frac{x}{2}+1]}^2\\ & =\frac{1}{2}{e}^2{(1+\tan \frac{x}{2})}^2\end{array}$ $\begin{array}{cc}& =\frac{1}{2}{e}^x[1+{\tan }^2\frac{x}{2}+2\tan \frac{x}{2}]\\ & =\frac{1}{2}{e}^x[1+{\tan }^2\frac{x}{2}+2\tan \frac{x}{2}]\\ & \frac{e^x(1+\sin x)dx}{(1+\cos x)}=e^x[\frac{1}{2}{\sec }^2\frac{x}{2}+\tan \frac{x}{2}]\end{array}$ Let $\tan \frac{x}{2}=f\left(x\right)f^{\prime }\left(x\right)=\frac{1}{2}{\sec }^2\frac{x}{2}$ It is known that $\int e^x\left\{f\right(x)+f^{\prime }(x\left)\right\}dx=e^{x}f\left(x\right)+C$ From equation (1) , we obtain $\int \frac{e^x(1+\sin x)}{(1+\cos x)}dx=e^x\tan \frac{x}{2}+C$

Q19. Integrate the function :$e^x(\frac{1}{x}-\frac{1}{x^2})$

Answer. Let $I=\int e^x[\frac{1}{x}-\frac{1}{x^2}]dx$ Also let $\frac{1}{x}=f\left(x\right)f^{\prime }\left(x\right)=\frac{-1}{x^2}$ It is known that, $\int e^x\left\{f\right(x)+f^{\prime }(x\left)\right\}dx=e^{x}f\left(x\right)+C$ $\therefore I=\frac{e^x}{x}+C$

Q20. Integrate the function :$\frac{(x-3)e^x}{(x-1{)}^3}$

Q21. Integrate the function :$e^{2x}\sin x$

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 Let $I=\int e^{2x}\sin xdx$ …(1) Integrating by parts , we obtain $\begin{array}{c}I=\sin x\int e^{2x}dx-\int \{\left(\frac{d}{dx}\sin x\right)\int e^{2x}dx\}dx\\ \Rightarrow I=\sin x\cdot \frac{e^{2x}}{2}-\int \cos x\cdot \frac{e^{2x}}{2}dx\\ \Rightarrow I=\frac{e^{2x}\sin x}{2}-\frac{1}{2}\int e^{2x}\cos xdx\end{array}$ Again integrating by parts, we obtain $\begin{array}{c}I=\frac{e^{2x}\cdot \sin x}{2}-\frac{1}{2}[\cos x\int e^{2x}dx-\int \{\left(\frac{d}{dx}\cos x\right)\int e^{2x}dx\}dx]\\ \Rightarrow I=\frac{e^{2x}\sin x}{2}-\frac{1}{2}[\cos x\cdot \frac{e^{2x}}{2}-\int (-\sin x\left)\frac{e^{2x}}{2}dx\right]\\ \Rightarrow I=\frac{e^{2x}\cdot \sin x}{2}-\frac{1}{2}[\frac{e^{2x}\cos x}{2}+\frac{1}{2}\int e^{2x}\sin xdx]\\ \Rightarrow I=\frac{e^{2x}\sin x}{2}-\frac{e^{2x}\cos x}{4}-\frac{1}{4}I\end{array}$ $\begin{array}{c}\Rightarrow I+\frac{1}{4}I=\frac{e^{2x}\cdot \sin x}{2}-\frac{e^{2x}\cos x}{4}\\ \Rightarrow \frac{5}{4}I=\frac{e^{2x}\sin x}{2}-\frac{e^{2x}\cos x}{4}\\ \Rightarrow I=\frac{4}{5}[\frac{e^{2x}\sin x}{2}-\frac{e^{2x}\cos x}{4}]+C\\ \Rightarrow I=\frac{e^{2x}}{5}[2\sin x-\cos x]+C\end{array}$

Q22. Integrate the function :${\sin }^{-1}\left(\frac{2x}{1+x^2}\right)$

Answer. Let $x=\tan \theta dx={\sec }^2\theta d\theta$ $\begin{array}{c}\therefore {\sin }^{-1}\left(\frac{2x}{1+x^2}\right)={\sin }^{-1}\left(\frac{2\tan \theta }{1+{\tan }^2\theta }\right)={\sin }^{-1}\left(\sin 2\theta \right)=2\theta \\ \int {\sin }^{-1}\left(\frac{2x}{1+x^2}\right)dx=\int 2\theta \cdot {\sec }^2\theta d\theta =2\int \theta \cdot {\sec }^2\theta d\theta \end{array}$ Integrating by parts, we obtain $\begin{array}{c}{}^2[\theta \cdot \int {\sec }^2\theta d\theta -\int \{\left(\frac{d}{d\theta }\theta \right)\int {\sec }^2\theta d\theta \}d\theta ]\\ =2[\theta \cdot \tan \theta -\int \tan \theta d\theta ]\\ =2[\theta \tan \theta +\log |\cos \theta ]+C\\ =2[x{\tan }^{-1}x+\log \left|\frac{1}{\sqrt{1+x^2}}\right|]+C\end{array}$ $\begin{array}{c}=2x{\tan }^{-1}x+2\log {(1+x^2)}^{\frac{1}{2}}+C\\ =2x{\tan }^{-1}x+2[-\frac{1}{2}\log (1+x^2)]+C\\ =2x{\tan }^{-1}x-\log (1+x^2)+C\end{array}$

Q23. Choose the correct answer : $\int x^2{e}^{x^3}dx$ equals (A) $\frac{1}{3}{e}^{x^3}+C$ (B) $\frac{1}{3}{e}^{x^2}+C$ (C) $\frac{1}{2}{e}^{x^3}+C$ (D) $\frac{1}{3}{e}^{x^2}+C$

Answer. Let $I=\int x^2{e}^{x^3}dx$ Also , let $x^3=t3x^{2}dx=dt$ $\begin{array}{cc}\Rightarrow I& =\frac{1}{3}\int e^{\prime }dt\\ & =\frac{1}{3}\left(e^t\right)+C\\ & =\frac{1}{3}{e}^{x^3}+C\end{array}$ Hence , the correct answer is A .

Q24. Choose the correct answer : $\int e^x\sec x(1+\tan x)dx$ equals (A) $e^x\cos x+C$ (B) $e^x\sec x+C$ (C) $e^x\sin x+C$ (D) $e^x\tan x+C$

Answer. $\int e^x\sec x(1+\tan x)dx$ Let $I=\int e^x\sec x(1+\tan x)dx=\int e^x(\sec x+\sec x\tan x)dx$ Also, let $\sec x=f\left(x\right)\sec x\tan x=f^{\prime }\left(x\right)$ It is known that , $\int e^x\left\{f\right(x)+f^{\prime }(x\left)\right\}dx=e^{x}f\left(x\right)+C$ $\therefore I=e^x\sec x+C$ Hence, the correct Answer is B.

Chapter-7 (integrals)

• Exercise 7.1
• Exercise 7.2
• Exercise 7.3
• Exercise 7.4
• Exercise 7.5
• Exercise 7.7
• Exercise 7.8
• Exercise 7.9
• Exercise 7.10
• Exercise 7.11