NCERT Solutions Class 12 maths Chapter-7 (Integrals)Exercise 7.2

NCERT Solutions Class 12 Maths from class 12th Students will get the answers of Chapter-7 (Integrals)Exercise 7.2 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 7.2

Q1. Integrate the function $\frac{2x}{1+x^2}$.

Answer.  $\begin{array}{c}\text{Let}1+x^2=t\\ \therefore 2xdx=dt\\ \Rightarrow \int \frac{2x}{1+x^2}dx=\int \frac{1}{1}dt\\ =\log \left|t\right|+C\\ =\log |1+x^2|+C\\ =\log (1+x^2)+C\end{array}$

Q2. Integrate the function $\frac{(\log x{)}^2}{x}$

Answer.  Let log |x| = t , $\therefore \frac{1}{x}dx=dt$ $\begin{array}{cc}\Rightarrow \int \frac{(\log |x|{)}^2}{x}dx& =\int t^{2}dt\\ & =\frac{t^3}{3}+C\\ & =\frac{(\log |x|{)}^3}{3}+C\end{array}$

Q3. Integrate the function$\frac{1}{x+x\log x}$

Answer. $\begin{array}{c}\frac{1}{x+x\log x}=\frac{1}{x(1+\log x)}\\ \text{Let}1+\log x=t\\ \frac{1}{x}dx=dt\end{array}$ $\begin{array}{cc}& \Rightarrow \int \frac{1}{x(1+\log x)}dx=\int \frac{1}{t}dt\\ & =\log |t|+C\\ & =\log |1+\log x|+C\end{array}$

Q4. Integrate the function: sin x ⋅ sin (cos x)

Answer. sin x ⋅ sin (cos x) Let cos x = t ∴ −sin x dx = dt $\begin{array}{cc}\Rightarrow \int \sin x\cdot \sin \left(\cos x\right)dx& =-\int \sin tdt\\ & =-[-\cos t]+C\\ & =\cos t+C\\ & =\cos \left(\cos x\right)+C\end{array}$

Q5. Integrate the function sin (ax + b ) cos ( ax + b )

Answer. $\sin (ax+b)\cos (ax+b)=\frac{2\sin (ax+b)\cos (ax+b)}{2}=\frac{\sin 2(ax+b)}{2}$ $\begin{array}{c}\text{Let}2(ax+b)=t\\ \therefore 2adx=dt\end{array}$ $\begin{array}{cc}\Rightarrow \int \frac{\sin 2(\alpha x+b)}{2}dx& =\frac{1}{2}\int \frac{\sin tdt}{2a}\\ & =\frac{1}{4a}[-\cos t]+C\\ & =\frac{-1}{4a}\cos 2(ax+b)+C\end{array}$

Q6. Integrate the function $\sqrt{ax+b}$

$\begin{array}{c}\text{Answer. Let}ax+b=t\\ \Rightarrow adx=dt\\ \therefore dx=\frac{1}{a}dt\end{array}$ $\begin{array}{cc}\Rightarrow & \int (ax+b{)}^{\frac{1}{2}}dx=\frac{1}{a}\int t^{\frac{1}{2}}dt\\ & =\frac{1}{a}\left(\frac{t^2}{\frac{3}{2}}\right)+C\\ & =\frac{2}{3a}(ax+b{)}^{\frac{3}{2}}+C\end{array}$

Q7. Integrate the function $x\sqrt{x+2}$

Answer. Let (x + 2) = t ∴ dx = dt $\begin{array}{cc}\Rightarrow \int x\sqrt{x+2}dx& =\int (t-2)\sqrt{t}dt\\ & =\int t^{\frac{3}{2}}-2t^{\frac{1}{2}})dt\\ & =\int t^{\frac{3}{2}}dt-2\int t^{\frac{1}{2}}dt\end{array}$ $\begin{array}{cc}& =\frac{t^2}{5}-2\left(\frac{t^{\frac{1}{2}}}{\frac{3}{2}}\right)+C\\ & =\frac{2}{5}{t}^{\frac{5}{2}}-\frac{4}{3}{t}^{\frac{3}{2}}+C\\ & =\frac{2}{5}(x+2{)}^{\frac{5}{2}}-\frac{4}{3}(x+2{)}^{\frac{3}{2}}+C\end{array}$

Q8. Integrate the function $x\sqrt{1+2x^2}$

Answer. $\begin{array}{c}\text{Let}1+2x^2=t\\ \therefore 4xdx=dt\end{array}$ $\begin{array}{cc}\Rightarrow \int x\sqrt{1+2x^2}dx& ={\int }_4^{\sqrt{t}dt}\\ & =\frac{1}{4}\int t^{\frac{1}{2}}dt\\ & =\frac{1}{4}\left(\frac{t^3}{2}\right)+C\\ & =\frac{1}{6}{(1+2x^2)}^{\frac{3}{2}}+C\end{array}$

Q9. Integrate the function: $(4x+2)\sqrt{x^2+x+1}$

Answer. $\begin{array}{c}\text{Let}{x}^2+x+1=t\\ \therefore (2x+1)dx=dt\\ \int (4x+2)\sqrt{x^2+x+1}dx\end{array}$ $\begin{array}{cc}& \int (4x+2)\sqrt{x^2+x+1}dx\\ & =\int 2\sqrt{t}dt\\ & =2\int \sqrt{t}dt\end{array}$ $\begin{array}{cc}& =2\left(\frac{t^{\frac{3}{2}}}{\frac{3}{2}}\right)+C\\ & =\frac{4}{3}{(x^2+x+1)}^{\frac{3}{2}}+C\end{array}$

Q10. Integrate the function : $\frac{1}{x-\sqrt{x}}$

Answer. $\begin{array}{c}\frac{1}{x-\sqrt{x}}=\frac{1}{\sqrt{x}(\sqrt{x}-1)}\\ \text{Let}(\sqrt{x}-1)=t\\ \frac{1}{2\sqrt{x}}dx=dt\end{array}$ $\begin{array}{cc}& \Rightarrow \int \frac{1}{\sqrt{x}(\sqrt{x}-1{)}^{dx}}=\int \frac{2}{t}dt\\ & =2\log |t|+C\\ & =2\log |\sqrt{x}-1|+C\end{array}$

Q11. Integrate the function : $\frac{x}{\sqrt{x+4}},x>0$

Answer. Let x + 4 = t ∴ dx = dt $\begin{array}{cc}\int \frac{x}{\sqrt{x+4}}dx& =\int \frac{(t-4)}{\sqrt{t}}dt\\ & =\int (\sqrt{t}-\frac{4}{\sqrt{t}})dt\\ & =\frac{t^{\frac{3}{2}}}{\frac{3}{2}}-4\left(\frac{\frac{1}{2}}{\frac{1}{2}}\right)+C\end{array}$ $\begin{array}{c}=\frac{2}{3}(t{)}^{\frac{3}{2}}-8(t{)}^{\frac{1}{2}}+C\\ =\frac{2}{3}t\cdot t^{\frac{1}{2}}-8t^{\frac{1}{2}}+C\\ =\frac{2}{3}{t}^{\frac{1}{2}}(t-12)+C\\ =\frac{2}{3}(x+4{)}^{\frac{1}{2}}(x+4-12)+C\\ =\frac{2}{3}\sqrt{x+4}(x-8)+C\end{array}$

Q12. Integrate the function: ${(x^3-1)}^{\frac{1}{3}}{x}^5$

Answer. $\begin{array}{c}\text{Let}{x}^3-1=t\\ \therefore 3x^{2}dx=dt\end{array}$ $\begin{array}{cc}\Rightarrow \int {(x^3-1)}^{\frac{1}{3}}{x}^{5}dx& =\int {(x^3-1)}^{\frac{1}{3}}{x}^3\cdot x^{2}dx\\ & =\int t^{\frac{1}{3}}(t+1)\frac{dt}{3}\\ & =\frac{1}{3}\int t^{\frac{4}{3}}+t^{\frac{1}{3}})dt\end{array}$ $\begin{array}{c}=\frac{1}{3}[\frac{t^{\frac{7}{3}}}{3}+\frac{t^{\frac{4}{3}}}{\frac{4}{3}}]+C\\ =\frac{1}{3}[\frac{3}{7}{t}^{\frac{1}{3}}+\frac{3}{4}{t}^{\frac{4}{3}}]+C\\ =\frac{1}{7}{(x^3-1)}^{\frac{7}{3}}+\frac{1}{4}{(x^3-1)}^{\frac{4}{3}}+C\end{array}$

Q13. Integrate the function : $\frac{x^2}{{(2+3x^3)}^3}$

Answer. $\begin{array}{c}\text{Let}2+3x^3=t\\ \therefore 9x^{2}dx=dt\end{array}$ $\begin{array}{cc}\Rightarrow \int \frac{x^2}{{(2+3x^3)}^3}dx& =\frac{1}{9}\int \frac{dt}{(t{)}^3}\\ & =\frac{1}{9}\left[\frac{t^{-2}}{-2}\right]+C\\ & =\frac{-1}{18}\left(\frac{1}{t^2}\right)+C\\ & =\frac{-1}{18{(2+3x^3)}^2}+C\end{array}$

Q14. Integrate the function : $\frac{1}{x(\log x{)}^m},x>0,m\ne 1$

Answer. Let log x = t $\begin{array}{c}\therefore \frac{1}{x}dx=dt\\ \Rightarrow \int \frac{1}{x(\log x{)}^m}dx=\int \frac{dt}{(t{)}^m}\\ =\frac{(\log x{)}^{1-\pi }}{(1-m)}+C\\ =\frac{(\log x{)}^{1-\pi }}{(1-m)}+C\end{array}$

Q15. Integrate the function : $\frac{x}{9-4x^2}$

Answer. $\begin{array}{c}\text{Let}9-4x^2=t\\ \therefore -8xdx=dt\\ \Rightarrow \int \frac{x}{9-4x^2}dx=\frac{-1}{8}\int \frac{1}{t}dt\end{array}$ $\begin{array}{c}=\frac{-1}{8}\log |t|+C\\ =\frac{-1}{8}\log |9-4x^2|+C\end{array}$

Q16. Integrate the function : $e^{2x+3}$

Answer. Let 2x + 3 = t ∴ 2dx = dt $\begin{array}{cc}\Rightarrow \int e^{2x+3}dx& =\frac{1}{2}\int e^{\prime }dt\\ & =\frac{1}{2}\left(e^t\right)+C\\ & =\frac{1}{2}{e}^{(2x+3)}+C\end{array}$

Q17. Integrate the function : $\frac{x}{e^{x^2}}$

Answer. $\begin{array}{c}\text{Let}{x}^2=t\\ \therefore 2xdx=dt\\ \Rightarrow \int \frac{x}{e^{x^2}}dx=\frac{1}{2}\int \frac{1}{e^{\prime }}dt\end{array}$ $\begin{array}{c}=\frac{1}{2}\int e^{-t}dt\\ =\frac{1}{2}\left(\frac{e^{-t}}{-1}\right)+C\\ =-\frac{1}{2}{e}^{-x^2}+C\\ =\frac{-1}{2e^{x^2}}+C\end{array}$

Q18. Integrate the function : $\frac{e^{{\tan }^{-1}x}}{1+x^2}$

Answer. $\begin{array}{c}\text{Let}{\tan }^{-1}x=t\\ \frac{1}{1+x^2}dx=dt\end{array}$ $\begin{array}{c}\Rightarrow \int \frac{e^{{\tan }^{-1}x}}{1+x^2}dx=\int e^{t}dt\\ =e^t+C\\ =e^{\ln n^{-1}x}+C\end{array}$

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

Answer. $\frac{e^{2x}-1}{e^{2x}+1}$ Dividing numerator and denominator by $e^x$ , we obtain $\begin{array}{c}\frac{(e^{2x}-1)}{e^{2x}+1)}=\frac{e^x-e^{-x}}{e^x+e^{-x}}\\ \frac{e^x+1)}{e^x}\\ \text{Let}{e}^x+e^{-x}=t\\ (e^x-e^{-x})dx=dt\end{array}$ $\begin{array}{cc}& \Rightarrow \int \frac{e^{2x}-1}{e^{2x}+1}dx=\int \frac{e^x-e^{-x}}{e^x+e^{-x}}dx\\ & =\int \frac{dt}{t}\\ & =\log |t|+C\\ & =\log |e^x+e^{-x}|+C\end{array}$

Q20. Integrate the function : $\frac{e^{2x}-e^{-2x}}{e^{2x}+e^{-2x}}$

Answer. $\begin{array}{c}\text{Let}{e}^{2x}+e^{-2x}=t\\ \therefore (2e^{2x}-2e^{-2x})dx=dt\\ \Rightarrow 2(e^{2x}-e^{-2x})dx=dt\end{array}$ $\begin{array}{cc}\Rightarrow 2(e^{2x}-e^{-2x})dx& =dt\\ \Rightarrow \int \left(\frac{e^{2x}-e^{-2x}}{e^{2x}+e^{-2x}}\right)dx& =\int \frac{dt}{2t}\\ & =\frac{1}{2}{\int }_t^{1}dt\\ & =\frac{1}{2}\log |t|+C\\ & =\frac{1}{2}\log |e^{2x}+e^{-2x}|+C\end{array}$

Q21. Integrate the function : ${\tan }^2(2x-3)$

Answer. $\begin{array}{c}{\tan }^2(2x-3)={\sec }^2(2x-3)-1\\ \text{Let}2x-3=t\\ \therefore 2dx=dt\end{array}$ $\begin{array}{cc}\Rightarrow \int {\tan }^2(2x-3)dx& =\int \left({\sec }^2\right(2x-3\left)\right)-1dx\\ & =\frac{1}{2}\int \left({\sec }^{2}t\right)dt-\int 1dx\\ & =\frac{1}{2}\int {\sec }^{2}tdt-\int \mathrm{ld}x\\ & =\frac{1}{2}\tan t-x+C\\ & =\frac{1}{2}\tan (2x-3)-x+C\end{array}$

Q22. Integrate the function : ${\sec }^2(7-4x)$

Answer. Let 7 − 4x = t ∴ −4dx = dt $\begin{array}{cc}\therefore \int {\sec }^2(7-4x)dx& =\frac{-1}{4}\int {\sec }^{2}tdt\\ & =\frac{-1}{4}\left(\tan t\right)+C\\ & =\frac{-1}{4}\tan (7-4x)+C\end{array}$

Q23. Integrate the function : $\frac{{\sin }^{-1}x}{\sqrt{1-x^2}}$

Answer. Let ${\sin }^{-1}x=t$ $\begin{array}{c}\therefore \frac{1}{\sqrt{1-x^2}}dx=dt\\ \Rightarrow \int \frac{{\sin }^{-1}x}{\sqrt{1-x^2}}dx=\int tdt\\ =\frac{{\left({\sin }^{-1}x\right)}^2}{2}+C\\ =\frac{{\left({\sin }^{-1}x\right)}^2}{2}+C\end{array}$

Q24. Integrate the function : $\frac{2\cos x-3\sin x}{6\cos x+4\sin x}$

Answer. $\begin{array}{c}\frac{2\cos x-3\sin x}{6\cos x+4\sin x}=\frac{2\cos x-3\sin x}{2(3\cos x+2\sin x)}\\ \text{Let}3\cos x+2\sin x=t\end{array}$ $\begin{array}{cc}& \\ (-3\sin x+2\cos x)dx=dt& \\ \int \frac{2\cos x-3\sin x}{6\cos x+4\sin x}dx& =\int \frac{dt}{2t}\\ & =\frac{1}{2}{\int }_t^{1}dt\\ & =\frac{1}{2}\log |t|+C\\ & =\frac{1}{2}\log |2\sin x+3\cos x|+C\end{array}$

Q25. Integrate the function : $\frac{1}{{\cos }^{2}x(1-\tan x{)}^2}$

Answer. $\begin{array}{c}\frac{1}{{\cos }^{2}x(1-\tan x{)}^2}=\frac{{\sec }^{2}x}{(1-\tan x{)}^2}\\ \mathrm{Let}(1-\tan x)=t\\ z-{\sec }^{2}xdx=dt\end{array}$ $\begin{array}{cc}\Rightarrow \int \frac{{\sec }^{2}x}{(1-\tan x{)}^2}dx& =\int \frac{-dt}{t^2}\\ & =-\int t^{-2}dt\\ & =+\frac{1}{t}+C\\ & =\frac{1}{(1-\tan x)}+C\end{array}$

Q26. Integrate the function : $\frac{\cos \sqrt{x}}{\sqrt{x}}$

Answer. $\begin{array}{c}\text{Let}\sqrt{x}=t\\ \frac{1}{2\sqrt{x}}dx=dt\end{array}$ $\begin{array}{cc}\Rightarrow \frac{\cos \sqrt{x}}{\sqrt{x}}dx& =2\int \cos tdt\\ & =2\sin t+C\\ & =2\sin \sqrt{x}+C\end{array}$

Q27. Integrate the function : $\sqrt{\sin 2x}\cos 2x$

Answer. $\begin{array}{c}\text{Let}\sin 2x=t\\ \therefore 2\cos 2xdx=dt\\ \Rightarrow \int \sqrt{\sin 2x}\cos 2xdx=\frac{1}{2}\int \sqrt{t}dt\end{array}$ $\begin{array}{c}=\frac{1}{2}\left(\frac{t^{\frac{3}{2}}}{\frac{3}{2}}\right)+C\\ =\frac{1}{3}{t}^{\frac{3}{2}}+C\\ =\frac{1}{3}(\sin 2x{)}^{\frac{3}{2}}+C\end{array}$

Q28. Integrate the function : $\frac{\cos x}{\sqrt{1+\sin x}}$

Answer. Let 1 + sin x = t $\begin{array}{cc}\therefore \cos xdx=dt& \\ \Rightarrow \int \frac{\cos x}{\sqrt{1+\sin x}}dx& =\int \frac{dt}{\sqrt{t}}\\ \Rightarrow & =\frac{t^{\frac{1}{2}}}{2}+C\\ & =2\sqrt{t}+C\\ & =2\sqrt{1+\sin x}+C\end{array}$

Q29. Integrate the function : cot x log sin x

Answer. Let log sin x = t $\begin{array}{cc}\Rightarrow \frac{1}{\sin x}\cdot \cos xdx& =dt\\ \therefore \cot xdx=dt& \\ \Rightarrow \cot x\log \sin xdx& =\int tdt\\ \Rightarrow & =\frac{t^2}{2}+C\\ & =\frac{1}{2}(\log \sin x{)}^2+C\end{array}$

Q30. Integrate the function : $\frac{\sin x}{1+\cos x}$

Answer. Let 1 + cos x = t ∴ −sin x dx = dt $\begin{array}{cc}\Rightarrow \int \frac{\sin x}{1+\cos x}dx& =\int -\frac{dt}{t}\\ & =-\log |t|+C\\ & =-\log |1+\cos x|+C\end{array}$

Q31. Integrate the function : $\frac{\sin x}{(1+\cos x{)}^2}$

Answer. Let 1 + cos x = t ∴ −sin x dx = dt $\begin{array}{cc}\Rightarrow \int \frac{\sin x}{(1+\cos x{)}^2}dx& =\int -\frac{dt}{t^2}\\ & =-\int t^{-2}dt\\ & =\frac{1}{t}+C\\ & =\frac{1}{1+\cos x}+C\end{array}$

Q32. Integrate the function : $\frac{1}{1+\cot x}$

Answer. $\begin{array}{cc}\text{Let}I& =\int \frac{1}{1+\cot x}dx\\ & =\int \frac{1}{1+\frac{\cos x}{\sin x}}dx\\ & =\int \frac{\sin x}{\sin x+\cos x}dx\\ & =\frac{1}{2}\int \frac{2\sin x}{\sin x+\cos x}dx\end{array}$ $\begin{array}{cc}& =\frac{1}{2}\int \frac{(\sin x+\cos x)+(\sin x-\cos x)}{(\sin x+\cos x)}dx\\ & =\frac{1}{2}\int 1dx+\frac{1}{2}\int \frac{\sin x-\cos x}{\sin x+\cos x}dx\\ & =\frac{1}{2}\left(x\right)+\frac{1}{2}\int \frac{\sin x-\cos x}{\sin x+\cos x}dx\end{array}$ Let sin x + cos x = t ⇒ (cos x − sin x) dx = dt $\begin{array}{cc}\therefore I& =\frac{x}{2}+\frac{1}{2}\int \frac{-\left(dt\right)}{t}\\ & =\frac{x}{2}-\frac{1}{2}\log |t|+C\\ & =\frac{x}{2}-\frac{1}{2}|\log \sin x+\cos x|+C\end{array}$

Q33. Integrate the function : $\frac{1}{1-\tan x}$

Answer. $\begin{array}{cc}\text{Let}I& =\int \frac{1}{1-\tan x}dx\\ & =\int \frac{1}{1-\frac{\sin x}{\cos x}}dx\\ & =\int \frac{\cos x}{\cos x-\sin x}dx\\ & =\frac{1}{2}\int \frac{2\cos x}{\cos x-\sin x}dx\end{array}$ $\begin{array}{cc}& =\frac{1}{2}\int \frac{(\cos x-\sin x)+(\cos x+\sin x)}{(\cos x-\sin x)}dx\\ & =\frac{1}{2}\int 1dx+\frac{1}{2}\int \frac{\cos x+\sin x}{\cos x-\sin x}dx\\ & =\frac{x}{2}+\frac{1}{2}\int \frac{\cos x+\sin x}{\cos x-\sin x}dx\end{array}$ Put cos x − sin x = t ⇒ (−sin x − cos x) dx = dt $\begin{array}{cc}\therefore I& =\frac{x}{2}+\frac{1}{2}\int \frac{-\left(dt\right)}{t}\\ & =\frac{x}{2}-\frac{1}{2}\log |t|+C\\ & =\frac{x}{2}-\frac{1}{2}\log |\cos x-\sin x|+C\end{array}$

Q34. Integrate the function : $\frac{\sqrt{\tan x}}{\sin x\cos x}$

Answer. $\begin{array}{cc}\text{Let}I& =\int \frac{\sqrt{\tan x}}{\sin x\cos x}dx\\ & =\int \frac{\sqrt{\tan x}x\cos x}{\sin x\cos x\times \cos x}dx\\ & =\int \frac{\sqrt{\tan x}}{\tan x{\cos }^{2}x}dx\\ & =\int \frac{{\sec }^{2}xdx}{\sqrt{\tan x}}\end{array}$ $\begin{array}{cc}\text{Let}& \tan x=t\Rightarrow {\sec }^{2}xdx=dt\\ \therefore I& =\int \frac{dt}{\sqrt{t}}\\ & =2\sqrt{t}+C\\ & =2\sqrt{\tan x}+C\end{array}$

Q35. Integrate the function : $\frac{(1+\log x{)}^2}{x}$

Answer. $\begin{array}{c}\frac{(x+1)(x+\log x{)}^2}{x}=\left(\frac{x+1}{x}\right)(x+\log x{)}^2=(1+\frac{1}{x})(x+\log x{)}^2\\ \text{Let}(x+\log x)=t\\ \dots (1+\frac{1}{x})dx=dt\end{array}$ $\begin{array}{cc}\Rightarrow (1+\frac{1}{x})(x+\log x{)}^{2}dx& =\int t^{2}dt\\ & =\frac{t^3}{3}+C\\ & =\frac{1}{3}(x+\log x{)}^3+C\end{array}$

Q36. Integrate the function : $\frac{(x+1)(x+\log x{)}^2}{x}$

Answer. $\begin{array}{c}\frac{(x+1)(x+\log x{)}^2}{x}=\left(\frac{x+1}{x}\right)(x+\log x{)}^2=(1+\frac{1}{x})(x+\log x{)}^2\\ \text{Let}(x+\log x)=t\\ \dots (1+\frac{1}{x})dx=dt\end{array}$ $\begin{array}{cc}\Rightarrow (1+\frac{1}{x})(x+\log x{)}^{2}dx& =\int t^{2}dt\\ & =\frac{t^3}{3}+C\\ & =\frac{1}{3}(x+\log x{)}^3+C\end{array}$

Q37. Integrate the function : $\frac{x^3\sin \left({\tan }^{-1}{x}^4\right)}{1+x^8}$

Answer. $\begin{array}{c}\text{Let}{x}^4=t\\ \therefore 4x^{3}dx=dt\end{array}$ $\begin{array}{c}\Rightarrow \int \frac{x^3\sin \left({\tan }^{-1}{x}^4\right)}{1+x^8}dx=\frac{1}{4}\int \frac{\sin \left({\tan }^{-1}t\right)}{1+t^2}dt....\left(1\right)\\ \text{Let}{\tan }^{-1}t=u\\ \frac{1}{1+t^2}dt=du\end{array}$ $\begin{array}{cc}\text{From}\left(1\right),\text{we obtain}& \\ \int \frac{x^3\sin \left({\tan }^{-1}{x}^4\right)dx}{1+x^8}=\frac{1}{4}\int \sin udu& \\ =& \frac{1}{4}(-\cos u)+C\end{array}$ $\begin{array}{cc}& =\frac{-1}{4}\cos \left({\tan }^{-1}t\right)+C\\ & =\frac{-1}{4}\cos \left({\tan }^{-1}{x}^4\right)+C\end{array}$

Q38. Choose the correct answer : $\int \frac{10x^9+{10}^x{\log }_{e}10dx}{x^{10}+{10}^x}$ equals $\begin{array}{cc}\text{(A)}{10}^x-x^{10}+C& \text{(B)}{10}^x+x^{10}+C\\ \text{(C)}{({10}^x-x^{10})}^{-1}+C& \text{(D)}\log ({10}^x+x^{10})+C\end{array}$

Answer. $\begin{array}{c}\text{Let}{x}^{10}+{10}^x=t\\ \therefore (10x^9+{10}^x{\log }_{e}10)dx=dt\end{array}$ $\begin{array}{c}\Rightarrow \int \frac{10x^9+{10}^x{\log }_{e}10}{x^{10}+{10}^x}dx=\int \frac{dt}{t}\\ =\log t+C\\ =\log ({10}^x+x^{10})+C\\ \text{Hence, the correct Answer is D.}\end{array}$

Q39. Choose the correct answer : $\int \frac{dx}{{\sin }^{2}x{\cos }^{2}x}$ equals $\begin{array}{cc}\text{(A)}\tan x+\cot x+C& \text{(B)}\tan x-\cot x+C\\ \text{(C)}\tan x\cot x+C& \text{(D)}\tan x-\cot 2x+C\end{array}$

Answer. $\begin{array}{cc}\text{Let}I& =\int \frac{dx}{{\sin }^{2}x{\cos }^{2}x}\\ & =\int \frac{1}{{\sin }^{2}x{\cos }^{2}x}dx\\ & =\int \frac{{\sin }^{2}x+{\cos }^{2}x}{{\sin }^{2}x{\cos }^{2}x}dx\end{array}$ $\begin{array}{cc}& =\int \frac{{\sin }^{2}x}{{\sin }^{2}x{\cos }^{2}x}dx+\int \frac{{\cos }^{2}x}{{\sin }^{2}x{\cos }^{2}x}dx\\ & =\int {\sec }^{2}xdx+\int {\csc }^{2}xdx\\ & =\tan x-\cot x+C\\ & \text{Hence, the correct Answer is}B\text{.}\end{array}$

Chapter-7 (integrals)

• Exercise 7.1
• Exercise 7.3
• Exercise 7.4
• Exercise 7.5
• Exercise 7.6
• Exercise 7.7
• Exercise 7.8
• Exercise 7.9
• Exercise 7.10
• Exercise 7.11