# 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 :

Answer. Let $I=\int x\mathrm{sin}xdx$ Taking x as first function and sin x as second function and integrating by parts, we obtain $I=x\int \mathrm{sin}xdx-\int \left\{\left(\frac{d}{dx}x\right)\int \mathrm{sin}xdx\right\}dx$

Q2. Integrate the function :

Answer. Let $I=\int x\mathrm{sin}3xdx$ Taking x as first function and sin 3x as second function and integrating by parts, we obtain $I=x\int \mathrm{sin}3xdx-\int \left\{\left(\frac{d}{dx}x\right)\int \mathrm{sin}3xdx\right\}$

Q3. Integrate the function :x2ex

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 $\begin{array}{rl}I& ={x}^{2}\int {e}^{x}dx-\int \left\{\left(\frac{d}{dx}{x}^{2}\right)\int {e}^{x}dx\right\}dx\\ & ={x}^{2}{e}^{x}-\int 2x\cdot {e}^{x}dx\\ & ={x}^{2}{e}^{x}-2\int x\cdot {e}^{x}dx\end{array}$ Again integrating by parts, we obtain

Q4. Integrate the function :

Answer. Let $I=\int x\mathrm{log}xdx$ Taking log x as first function and x as second function and integrating by parts, we obtain

Q5. Integrate the function :

Answer. Let $I=\int x\mathrm{log}2xdx$ Taking log 2x as first function and x as second function and integrating by parts, we obtain

Q6. Integrate the function :x2logx

Answer. Let $I=\int {x}^{2}\mathrm{log}xdx$ Taking log x as first function and ${x}^{2}$ as second function and integrating by parts, we obtain

Q7. Integrate the function :xsin1x

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

Q8. Integrate the function :xtan1x

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

Q9. Integrate the function :

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