Maths NCERT Exemplar Solutions Class 12th Chapter Seven: Overview, Questions, Preparation

Sol:

$\begin{array}{l}LetI={\displaystyle \int \frac{x^2}{\left(x^2+a^2\right)\left(x^2+b^2\right)}}dx\\ Put{x}^2=tforthepurposeofpartialfraction.\\ Weget\frac{t}{\left(t+a^2\right)\left(t+b^2\right)}\\ Put\frac{t}{\left(t+a^2\right)\left(t+b^2\right)}=\frac{A}{t+a^2}+\frac{B}{t+b^2}\left[whereAandBarearbitrar\mathrm{yconstants}.\right]\\ \frac{t}{\left(t+a^2\right)\left(t+b^2\right)}=\frac{A\left(t+b^2\right)+B\left(t+a^2\right)}{\left(t+a^2\right)\left(t+b^2\right)}\\ \Rightarrow t=At+A{b}^2+Bt+B{a}^2\\ Comparingtheliketerms,weget\\ A+B=1andA{b}^2+B{a}^2=0\\ \Rightarrow A=\frac{-a^2}{b^2}B\\ \therefore \frac{-a^2}{b^2}B+B=1\\ B\left(\frac{-a^2}{b^2}+1\right)=1\Rightarrow B\left(\frac{-a^2+b^2}{b^2}\right)=1\\ \Rightarrow B=\frac{b^2}{b^2-a^2}andA=\frac{-a^2}{b^2}\times \frac{b^2}{b^2-a^2}=\frac{a^2}{a^2-b^2}\\ So,A=\frac{a^2}{a^2-b^2}andB=\frac{-b^2}{a^2-b^2}\\ \therefore {\displaystyle \int \frac{x^2}{\left(x^2+a^2\right)\left(x^2+b^2\right)}}dx=\frac{a^2}{a^2-b^2}{\displaystyle \int \frac{1}{x^2+a^2}dx-}\frac{b^2}{a^2-b^2}{\displaystyle \int \frac{1}{x^2+b^2}dx}\\ =\frac{a^2}{a^2-b^2}\times \frac{1}{a}ta{n}^{-1}\frac{x}{a}-\frac{b^2}{a^2-b^2}.\frac{1}{b}.ta{n}^{-1}\frac{x}{b}\\ =\frac{a}{a^2-b^2}ta{n}^{-1}\frac{x}{a}-\frac{b}{a^2-b^2}ta{n}^{-1}\frac{x}{b}+C\\ Hence,I=\frac{1}{a^2-b^2}\left[ata{n}^{-1}\frac{x}{a}-bta{n}^{-1}\frac{x}{b}\right]+C.\end{array}$

Sol:

$\begin{array}{l}L.H.S.={\displaystyle \int \frac{2x-1}{2x+3}}dx\\ \Rightarrow {\displaystyle \int \left(1-\frac{4}{2x+3}\right)}dx\left[Dividingthenumeratorbythe\text{\hspace{0.17em}denominator}\right]\\ \Rightarrow {\displaystyle \int 1.}dx-4{\displaystyle \int \frac{1}{2x+3}dx}\Rightarrow {\displaystyle \int 1.}dx-\frac{4}{2}{\displaystyle \int \frac{1}{x+\frac{3}{2}}dx}\\ \Rightarrow {\displaystyle \int 1.}dx-2{\displaystyle \int \frac{1}{x+\frac{3}{2}}dx}\Rightarrow x-2log\left|x+\frac{3}{2}\right|+C\\ \Rightarrow x-2log\left|\frac{2x+3}{2}\right|+C\Rightarrow x-log\left|{\left(\frac{2x+3}{2}\right)}^2\right|+C\\ \left[\because nlogm=log{m}^n\right]\\ \Rightarrow x-log\left|{\left(2x+3\right)}^2\right|-log{2}^2+C\\ \Rightarrow x-log\left|{\left(2x+3\right)}^2\right|+C_1=R.H.S.\left[where{C}_1=C-log{2}^2\right]\\ L.H.S.=R.H.S.\\ Hence,proved.\end{array}$

Sol:

$\begin{array}{l}L.H.S.={\displaystyle \int \frac{2x+3}{x^2+3x}}dx\\ Put{x}^2+3x=t\\ \therefore \left(2x+3\right)dx=dt\\ \Rightarrow {\displaystyle \int \frac{dt}{t}}=log\left|t\right|\Rightarrow log\left|x^2+3x\right|+C=R.H.S.\\ L.H.S.=R.H.S.\\ Hence,proved.\end{array}$

Choose the correct option from given four options in each of the Exercises from 1 to 11.

Q1. $\int \frac{cos2x-cos2\theta }{cosx-cos\theta } dx$ is equal to

(A)  $2\left(sinx+xcos\theta \right)+C$

(B)  $2\left(sinx-xcos\theta \right)+C$

(C)  $2\left(sinx+2xcos\theta \right)+C$

(D)  $2\left(sinx-2xcos\theta \right)+C$

Sol:

$\begin{array}{l}LetI={\displaystyle \int \frac{cos2x-cos2\theta }{cosx-cos\theta } dx}\\ ={\displaystyle \int \frac{\left(2co{s}^{2}x-1\right)-\left(2co{s}^2\theta -1\right)}{cosx-cos\theta } dx}\\ ={\displaystyle \int \frac{2co{s}^{2}x-1-2co{s}^2\theta +1}{cosx-cos\theta } dx}\\ ={\displaystyle \int \frac{2co{s}^{2}x-2co{s}^2\theta }{cosx-cos\theta } dx}=2{\displaystyle \int \frac{co{s}^{2}x-co{s}^2\theta }{cosx-cos\theta } dx}\\ =2{\displaystyle \int \frac{\left(cosx+cos\theta \right)\left(cosx-cos\theta \right)}{\left(cosx-cos\theta \right)} dx}\\ =2{\displaystyle \int \left(cosx+cos\theta \right)dx=}2\left(sinx+cos\theta .x\right)+C\\ Hence,correctoptionis\left(a\right).\end{array}$

Q2. $\int \frac{dx}{sin\left(x-a\right)sin\left(x-b\right)} is equal to$

(A)  $sin\left(b-a\right)log\left| \frac{sin\left(x-a\right)}{sin\left(x-b\right)} \right|+C$

(B)  $cosec\left(b-a\right)log\left| \frac{sin\left(x-a\right)}{sin\left(x-b\right)} \right|+C$

(C)  $cosec\left(b-a\right)log\left| \frac{sin\left(x-a\right)}{sin\left(x-b\right)} \right|+C$

(D)  $sin\left(b-a\right)log\left| \frac{sin\left(x-a\right)}{sin\left(x-b\right)} \right|+C$

Sol:

$\begin{array}{l}LetI={\displaystyle \int \frac{dx}{sin\left(x-a\right).sin\left(x-b\right)} }\\ Multiplyinganddividingbysin\left(b-a\right)\\ =\frac{1}{sin\left(b-a\right)}{\displaystyle \int \frac{sin\left(b-a\right)}{sin\left(x-a\right).sin\left(x-b\right)} dx}\\ =\frac{1}{sin\left(b-a\right)}{\displaystyle \int \frac{sin\left(x+b-x-a\right)}{sin\left(x-a\right).sin\left(x-b\right)} dx}\\ =\frac{1}{sin\left(b-a\right)}{\displaystyle \int \frac{sin\left[\left(x-a\right)-\left(x-b\right)\right]}{sin\left(x-a\right).sin\left(x-b\right)} dx}\\ =\frac{1}{sin\left(b-a\right)}{\displaystyle \int \frac{sin\left[\left(x-a\right)-\left(x-b\right)\right]}{sin\left(x-a\right).sin\left(x-b\right)} dx}\\ =\frac{1}{sin\left(b-a\right)}{\displaystyle \int \frac{sin\left(x-a\right)cos\left(x-b\right)-cos\left(x-a\right)sin\left(x-b\right)}{sin\left(x-a\right).sin\left(x-b\right)} dx}\\ =\frac{1}{sin\left(b-a\right)}{\displaystyle \int \frac{sin\left(x-a\right)cos\left(x-b\right)}{sin\left(x-a\right).sin\left(x-b\right)}-\frac{cos\left(x-a\right)sin\left(x-b\right)}{sin\left(x-a\right).sin\left(x-b\right)} dx}\\ =\frac{1}{sin\left(b-a\right)}{\displaystyle \int \left[\frac{cos\left(x-b\right)}{sin\left(x-b\right)}-\frac{cos\left(x-a\right)}{sin\left(x-a\right)}\right]}dx\\ =\frac{1}{sin\left(b-a\right)}{\displaystyle \int \left[cot\left(x-b\right)-cot\left(x-a\right)\right]}dx\\ =\frac{1}{sin\left(b-a\right)}\left[logsin\left(x-b\right)-logsin\left(x-a\right)\right]+C\\ =\frac{1}{sin\left(b-a\right)}.log\left|\frac{sin\left(x-b\right)}{sin\left(x-a\right)}\right|+C\\ I=cosec\left(b-a\right).log\left|\frac{sin\left(x-b\right)}{sin\left(x-a\right)}\right|+C\\ Hence,correctoptionis\left(c\right).\end{array}$

Sol:

$\begin{array}{l}LetI={\displaystyle \underset{0}{\overset{\frac{\pi }{2}}{\int }}cosx.e^{sinx}}dx\\ Putsinx=t\Rightarrow cosxdx=dt\\ Whenx=0thent=sin0=0;Whenx=\frac{\pi }{2}thent=sin\frac{\pi }{2}=1\\ \therefore I={\displaystyle \underset{0}{\overset{1}{\int }}{e}^{t}dt}={\left[e^t\right]}_0^1=\left(e^1-e^0\right)=e-1\\ Hence,I=e-1\end{array}$

Sol:

$\begin{array}{l}LetI={\displaystyle \int \frac{x+3}{{\left(x+4\right)}^2}.e^x dx}\\ ={\displaystyle \int \frac{x+4-1}{{\left(x+4\right)}^2}.e^x dx}\\ ={\displaystyle \int \left[\frac{x+4}{{\left(x+4\right)}^2}-\frac{1}{{\left(x+4\right)}^2}\right].e^x dx}={\displaystyle \int \left[\frac{1}{x+4}-\frac{1}{{\left(x+4\right)}^2}\right].e^x dx}\\ Put\frac{1}{x+4}=t\Rightarrow -\frac{1}{{\left(x+4\right)}^2}dx=dt\\ Letf\left(x\right)=\frac{1}{x+4}\therefore f^{\text{'}}\left(x\right)=-\frac{1}{{\left(x+4\right)}^2}\\ Using{\displaystyle \int e^x\left[f\left(x\right)+f^{\text{'}}\left(x\right)\right]dx=}{e}^{x}f\left(x\right)+C\\ \therefore I=e^x.\frac{1}{x+4}+C\\ Hence,I=\frac{e^x}{x+4}+C.\end{array}$