NCERT Solutions Class 12 Maths Chapter-7 (Integrals)Exercise 7.10

NCERT Solutions Class 12 Maths from class 12th Students will get the answers of Chapter-7 (Integrals)Exercise 7.10 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 7.10

Q1. Evaluate the definite integral : ${\int }_0^1\frac{x}{x^2+1}dx$ using substitution

Answer. ${\int }_0^1\frac{x}{x^2+1}dx$ $\begin{array}{c}\text{Let}{x}^2+1=t\Rightarrow 2xdx=dt\\ \text{when}x=0,t=1\text{and when}x=1,t=2\\ \therefore {\int }_0^1\frac{x}{x^2+1}dx=\frac{1}{2}{\int }_1^2\frac{dt}{t}\end{array}$ $\begin{array}{c}=\frac{1}{2}[\log |t{]}_1^2\\ =\frac{1}{2}[\log 2-\log 1]\\ =\frac{1}{2}\log 2\end{array}$

Q2. Evaluate the definite integral : ${\int }_0^{\pi }\sqrt{\sin \varphi }{\cos }^5\varphi d\varphi$ using substitution

Answer. Let $I={\int }_0^{\frac{\pi }{2}}\sqrt{\sin \varphi {\cos }^5\varphi d\varphi }={\int }_b^{\pi }\sqrt{\sin \varphi }{\cos }^4\varphi \cos \varphi d\varphi$ Also, let $\sin \varphi =t\Rightarrow \cos \varphi d\varphi =dt$ $\begin{array}{cc}\text{When}\varphi & =0,t=0\text{and when}\varphi =\frac{\pi }{2},t=1\\ \therefore I& ={\int }_0^1\sqrt{t}{(1-t^2)}^{2}dt\\ & ={\int }_0^1{t}^{\frac{1}{2}}(1+t^4-2t^2)dt\end{array}$ $\begin{array}{cc}& ={\int }_0^1[t^{\frac{1}{2}}+t^{\frac{9}{2}}-2t^{\frac{5}{2}}]dt\\ & ={[\frac{t^{\frac{3}{2}}}{\frac{3}{2}}+\frac{t^{\frac{11}{2}}}{\frac{11}{2}}-\frac{2t^{\frac{7}{2}}}{\frac{7}{2}}]}_0^1\end{array}$ $\begin{array}{c}=\frac{2}{3}+\frac{2}{11}-\frac{4}{7}\\ =\frac{154+42-132}{231}\\ =\frac{64}{231}\end{array}$

Q3. Evaluate the definite integral : ${\int }_0^1{\sin }^{-1}\left(\frac{2x}{1+x^2}\right)dx$ using substitution

Answer. $\begin{array}{c}\text{Let}I={\int }_0^1{\sin }^{-1}\left(\frac{2x}{1+x^2}\right)dx\\ \text{Also, let}x=\tan \theta dx={\sec }^2\theta d\theta \\ \text{when}x=0,\theta =0\text{and when}x=1,\theta =\frac{\pi }{4}\end{array}$ $I={\int }_0^{\frac{\pi }{4}}{\sin }^{-1}\left(\frac{2\tan \theta }{1+{\tan }^2\theta }\right){\sec }^2\theta d\theta$ $\begin{array}{cc}& ={\int }_0^{\frac{\pi }{4}}{\sin }^{-1}\left(\sin 2\theta \right){\sec }^2\theta d\theta \\ & ={\int }_0^{\frac{\pi }{4}}2\theta \cdot {\sec }^2\theta d\theta \\ & =2{\int }_0^{\frac{\pi }{4}}\theta \cdot {\sec }^2\theta d\theta \end{array}$ $\begin{array}{c}\text{Taking}\theta \text{as first function and}{\sec }^2\theta \text{as second function and integrating by parts, we obtain}\\ I=2{[\theta \int \theta \int {\sec }^2\theta d\theta -\int \{\left(\frac{d}{dx}\theta \right)\int {\sec }^2\theta d\theta \}d\theta ]}_0^{\frac{\pi }{4}}\end{array}$ $\begin{array}{cc}I& =2{[\theta \int \theta \int {\sec }^2\theta d\theta -\int \{\left(\frac{d}{dx}\theta \right)\int {\sec }^2\theta d\theta \}d\theta ]}_0^{\frac{x}{4}}\\ & =2{[\theta \tan \theta -\int \tan \theta d\theta ]}_0^{\frac{\pi }{4}}\\ & =2[\theta \tan \theta +\log \cos \theta {]}_0^{\frac{\pi }{4}}\end{array}$ $\begin{array}{cc}& =2[\frac{\pi }{4}\tan \frac{\pi }{4}+\log \cos \frac{\pi }{4}|-\log |\cos 0|]\\ & =2[\frac{\pi }{4}+\log \left(\frac{1}{\sqrt{2}}\right)-\log 1]\\ & =2[\frac{\pi }{4}-\frac{1}{2}\log 2]\\ & =\frac{\pi }{2}-\log 2\end{array}$

Q4. Evaluate the definite integral : ${\int }_0^{2}x\sqrt{x+2}(\text{Put}x+2=t^2)$ using substitution

Answer. $\begin{array}{c}{\int }_0^{2}x\sqrt{x+2}dx\\ \text{Let}x+2=t^{2}dx=2tdt\\ \text{when}x=0,t=\sqrt{2}\text{and when}x=2,t=2\end{array}$ $\begin{array}{cc}\therefore {\int }_0^{2}x\sqrt{x+2}dx& ={\int }_{\sqrt{2}}^2(t^2-2)\sqrt{t^2}2tdt\\ & =2{\int }_{\sqrt{2}}^2{(t^2-2)}^{2}dt\\ & =2{\int }_{\sqrt{2}}^2(t^4-2t^2)dt\\ & =2{[\frac{t^5}{5}-\frac{2t^3}{3}]}_{\sqrt{2}}^2\end{array}$ $\begin{array}{cc}& =2[\frac{32}{5}-\frac{16}{3}-\frac{4\sqrt{2}}{5}+\frac{4\sqrt{2}}{3}]\\ & =2\left[\frac{96-80-12\sqrt{2}+20\sqrt{2}}{15}\right]\\ & =2\left[\frac{16+8\sqrt{2}}{15}\right]\\ & =\frac{16(2+\sqrt{2})}{15}\\ & =\frac{16(2+\sqrt{2})}{15}\\ & =\frac{16\sqrt{2}(\sqrt{2}+1)}{15}\end{array}$

Q5. Evaluate the definite integral : ${\int }_0^{\frac{\pi }{2}}\frac{\sin x}{1+{\cos }^{2}x}dx$ using substitution

Answer. $\begin{array}{c}{\int }_0^{\frac{\pi }{2}}\frac{\sin x}{1+{\cos }^{2}x}dx\\ \text{Let}\cos x=t-\sin xdx=dt\\ \text{when}x=0,t=1\text{and when}x=\frac{\pi }{2},t=0\end{array}$ $\begin{array}{cc}\Rightarrow {\int }_0^{\frac{\pi }{2}}\frac{\sin x}{1+{\cos }^{2}x}dx& =-\int \frac{dt}{1+t^2}\\ & =-{\left[{\tan }^{-1}t\right]}_1^0\\ & =-[{\tan }^{-1}0-{\tan }^{-1}1]\\ & =-[-\frac{\pi }{4}]\\ & =\frac{\pi }{4}\end{array}$

Q6. Evaluate the definite integral : ${\int }_0^2\frac{dx}{x+4-x^2}$ using substitution

Answer. $\begin{array}{cc}{\int }_0^2\frac{dx}{x+4-x^2}& ={\int }_0^2\frac{dx}{-(x^2-x-4)}\\ & ={\int }_0^2\frac{dx}{-(x^2-x+\frac{1}{4}-\frac{1}{4}-4})\end{array}$ $\begin{array}{cc}& ={\int }_0^2\frac{dx}{-[{(x-\frac{1}{2})}^2-\frac{17}{4}]}\\ & ={\int }_0^2\frac{dx}{{\left(\frac{\sqrt{17}}{2}\right)}^2-{(x-\frac{1}{2})}^2}\end{array}$ Let $x-\frac{1}{2}=tdx=dt$ $\begin{array}{c}\text{When}x=0,t=-\frac{1}{2}\text{and when}x=2,t=\frac{3}{2}\\ \therefore {\int }_0^2\frac{dx}{{\left(\frac{\sqrt{17}}{2}\right)}^2-{(x-\frac{1}{2})}^2}={\int }_{-\frac{1}{2}}^{\frac{3}{2}}\frac{dt}{{\left(\frac{\sqrt{17}}{2}\right)}^2-t^2}\end{array}$ $={\left[\frac{1}{2\left(\frac{\sqrt{17}}{2}\right)}\log \frac{\frac{\sqrt{17}}{2}+t}{\frac{\sqrt{17}}{2}-t}\right]}_{\frac{1}{2}}^{\frac{3}{2}}$ $\begin{array}{cc}& =\frac{1}{\sqrt{17}}[\log \frac{\frac{\sqrt{17}}{2}+\frac{3}{2}}{\frac{\sqrt{17}}{2}-\frac{3}{2}}-\frac{\log \frac{\sqrt{17}}{2}-\frac{1}{2}}{\log \frac{\sqrt{17}}{2}+\frac{1}{2}}]\\ & =\frac{1}{\sqrt{17}}[\log \frac{\sqrt{17}+3}{\sqrt{17}-3}-\log \frac{\sqrt{17}-1}{\sqrt{17}+1}]\end{array}$ $\begin{array}{cc}& =\frac{1}{\sqrt{17}}\log \frac{\sqrt{17}+3}{\sqrt{17}-3}\times \frac{\sqrt{17}+1}{\sqrt{17}-1}\\ & =\frac{1}{\sqrt{17}}\log \left[\frac{17+3+4\sqrt{17}}{17+3-4\sqrt{17}}\right]\end{array}$ $\begin{array}{cc}& =\frac{1}{\sqrt{17}}\log \left[\frac{20+4\sqrt{17}}{20-4\sqrt{17}}\right]\\ & =\frac{1}{\sqrt{17}}\log \left(\frac{5+\sqrt{17}}{5-\sqrt{17}}\right)\end{array}$ $\begin{array}{c}=\frac{1}{\sqrt{17}}\log \left[\frac{(5+\sqrt{17})(5+\sqrt{17})}{25-17}\right]\\ =\frac{1}{\sqrt{17}}\log \left[\frac{25+17+10\sqrt{17}}{8}\right]\end{array}]$ $\begin{array}{cc}& =\frac{1}{\sqrt{17}}\log \left(\frac{42+10\sqrt{17}}{8}\right)\\ & =\frac{1}{\sqrt{17}}\log \left(\frac{21+5\sqrt{17}}{4}\right)\end{array}$

Q7. Evaluate the definite integral : ${\int }_{-1}^1\frac{dx}{x^2+2x+5}$ using substitution

Answer. ${\int }_{-1}^1\frac{dx}{x^2+2x+5}={\int }_{-1}^1\frac{dx}{(x^2+2x+1)+4}={\int }_{-1}^2\frac{dx}{(x+1{)}^2+(2{)}^2}$ Let x + 1 = t ⇒ dx = dt When x = −1, t = 0 and when x = 1, t = 2 $\therefore {\int }_{-1}^4\frac{dx}{(x+1{)}^2+(2{)}^2}={\int }_3^2\frac{dt}{t^2+2^2}$ $\begin{array}{c}={\left[\frac{1}{2}{\tan }^{-1}\frac{t}{2}\right]}_0^2\\ =\frac{1}{2}{\tan }^{-1}1-\frac{1}{2}{\tan }^{-1}0\\ =\frac{1}{2}\left(\frac{\pi }{4}\right)=\frac{\pi }{8}\end{array}$

Q8. Evaluate the definite integral : ${\int }_1^2(\frac{1}{x}-\frac{1}{2x^2})e^{2x}dx$ using substitution

Answer. $\begin{array}{c}{\int }_1^2(\frac{1}{x}-\frac{1}{2x^2})e^{2x}dx\\ \text{Let}2x=t2dx=dt\\ \text{When}x=1,t=2\text{and when}x=2,t=4\end{array}$ $\begin{array}{cc}\therefore {\int }_1^2(\frac{1}{x}-\frac{1}{2x^2})e^{2x}dx& =\frac{1}{2}{\int }_2^1(\frac{2}{t}-\frac{2}{t^2})e^{t}dt\\ & ={\int }_2^4(\frac{1}{t}-\frac{1}{t^2})e^{t}dt\\ \text{Let}\frac{1}{t}=f\left(t\right)& \end{array}$ $\text{Then,}{f}^{\prime }\left(t\right)=-\frac{1}{t^2}$ $\begin{array}{cc}\Rightarrow {\int }_2^1(\frac{1}{t}-\frac{1}{t^2})e^{t}dt& ={\int }_2^t{e}^t\left[f\right(t)+f^t(t\left)\right]dt\\ & ={\left[e^{t}f\right(t\left)\right]}_2^4\end{array}$ $\begin{array}{c}={[e^t\cdot \frac{2}{t}]}_2^4\\ ={\left[\frac{e^t}{t}\right]}_2^4\\ =\frac{e^4}{4}-\frac{e^2}{2}\\ =\frac{e^2(e^2-2)}{4}\end{array}$

Q9. The value of the integral ${\int }_{\frac{1}{3}}\frac{{(x-x^3)}^{\frac{1}{3}}}{x^4}dx$ is (A) 6 (B) 0 (C) 3 (D) 4

Answer. Let $I={\int }_{\frac{1}{3}}^{{(x-x^3)}^{\frac{1}{3}}}dx$ Also, let $x=\sin \theta \Rightarrow dx=\cos \theta d\theta$ When $x=\frac{1}{3},\theta ={\sin }^{-1}\left(\frac{1}{3}\right)\text{and when}x=1,\theta =\frac{\pi }{2}$ $\Rightarrow I={\int }_{{lim}^{-1}\left(\frac{1}{3}\right)}^{\frac{\pi }{2}}\frac{{(\sin \theta -{\sin }^3\theta )}^{\frac{1}{3}}}{{\sin }^4\theta }\cos \theta d\theta$ $={\int }_{-{\sin }^{-1}\left(\frac{1}{3}\right)}^{\frac{x}{2}}\frac{(\sin \theta {)}^{\frac{1}{3}}{(1-{\sin }^2\theta )}^{\frac{1}{3}}}{{\sin }^4\theta }\cos \theta d\theta$ $={\int }_{-\pi }^{\frac{\pi }{2}}\frac{\left(\sin \theta {)}^{\frac{1}{3}}\right(\cos \theta {)}^{\frac{2}{3}}}{{\sin }^4\theta }\cos \theta d\theta$ $={\int }_{-{\ln }^{-1}\left(\frac{1}{3}\right)}^{\frac{\pi }{2}}\frac{\left(\sin \theta {)}^{\frac{1}{3}}\right(\cos \theta {)}^{\frac{2}{3}}}{{\sin }^2\theta {\sin }^2\theta }\cos \theta d\theta$ $={\int }_{{\mathrm{lin}}^{-1}\left(\frac{1}{3}\right)}^{\frac{\pi }{2}}\frac{(\cos \theta {)}^{\frac{5}{3}}}{(\sin \theta {)}^{\frac{5}{3}}}{\csc }^2\theta d\theta$ $={\int }_{{\ln }^{-1}\left(\frac{1}{3}\right)}^{\frac{\pi }{2}}\frac{(\cos \theta {)}^{\frac{5}{3}}}{(\sin \theta {)}^{\frac{5}{3}}}{\csc }^2\theta d\theta$ $={\int }_{-{\ln }^{-1}\left(\frac{1}{3}\right)}^{\frac{x}{2}}(\cot \theta {)}^{\frac{5}{3}}{\csc }^2\theta d\theta$ $\cot \theta =t-\csc 2\theta d\theta =dt$ When $\theta ={\sin }^{-1}\left(\frac{1}{3}\right),t=2\sqrt{2}\text{and when}\theta =\frac{\pi }{2},t=0$ $\begin{array}{cc}\therefore I& =-{\int }_{\sqrt{2}}^0(t{)}^{\frac{5}{3}}dt\\ & =-{\left[\frac{3}{8}\right(t{)}^{\frac{8}{3}}]}_{2\sqrt{2}}^0\\ & =-\frac{3}{8}{\left[\right(t{)}^{\frac{8}{3}}]}_{2\sqrt{2}}^9\\ & =-\frac{3}{8}[-(2\sqrt{2}{)}^{\frac{8}{3}}]\end{array}$ $\begin{array}{cc}& =\frac{3}{8}\left[\right(\sqrt{8}{)}^{\frac{8}{3}}]\\ & =\frac{3}{8}\left[\right(8{)}^{\frac{4}{3}}]\\ & =\frac{3}{8}\left[16\right]\\ & =3\times 2\\ & =6\end{array}$ Hence, the correct Answer is A.

Q10. If $f\left(x\right)={\int }_0^{\pi }t\sin tdt,\text{then}{f}^{\prime }\left(x\right)$ is A. $\cos x+x\sin x$ B. $x\sin x$ C. $x\cos x$ D. $\sin x+x\cos x$

Answer. $f\left(x\right)={\int }_0^{x}t\sin tdt$ Integrating by parts, we obtain $\begin{array}{cc}f\left(x\right)& =t{\int }_b^{\pi }\sin tdt-{\int }_0^r\{\left(\frac{d}{dt}t\right)\int \sin tdt\}dt\\ & =\left[t\right(-\cos t){]}_0^x-{\int }_0^x(-\cos t)dt\\ & =[-t\cos t+\sin t{]}_0^x\\ & =-x\cos x+\sin x\end{array}$ $\begin{array}{cc}\Rightarrow f^{\prime }\left(x\right)& =-\left[\right\{x(-\sin x)\}+\cos x]+\cos x\\ & =x\sin x-\cos x+\cos x\\ & =x\sin x\end{array}$ 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 7.6
• Exercise 7.7
• Exercise 7.8
• Exercise 7.9
• Exercise 7.11