Class 12th

This is a Short answer type Questions as classified in NCERT Exemplar

Sol:

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

This is a Short answer type Questions as classified in NCERT Exemplar

Sol:

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

This is a Long answer type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l}LetI={\displaystyle \underset{-\frac{\pi }{4}}{\overset{\frac{\pi }{4}}{\int }}log\left|sinx+cosx\right|dx}\dots \left(i\right)\\ ={\displaystyle \underset{-\frac{\pi }{4}}{\overset{\frac{\pi }{4}}{\int }}log\left|sin\left(\frac{\pi }{4}-\frac{\pi }{4}-x\right)+cos\left(\frac{\pi }{4}-\frac{\pi }{4}-x\right)\right|dx}\left[Using{\displaystyle \underset{a}{\overset{b}{\int }}f\left(x\right)dx={\displaystyle \underset{a}{\overset{b}{\int }}f\left(a+b-x\right)dx}}\right]\\ ={\displaystyle \underset{-\frac{\pi }{4}}{\overset{\frac{\pi }{4}}{\int }}log\left|sin\left(-x\right)+cosx\right|dx}\\ ={\displaystyle \underset{-\frac{\pi }{4}}{\overset{\frac{\pi }{4}}{\int }}log\left|cosx-sinx\right|dx}\dots \left(ii\right)\\ Adding\left(i\right)and\left(ii\right)\\ 2I={\displaystyle \underset{-\frac{\pi }{4}}{\overset{\frac{\pi }{4}}{\int }}log\left|cosx+sinx\right|dx}+{\displaystyle \underset{-\frac{\pi }{4}}{\overset{\frac{\pi }{4}}{\int }}log\left|cosx-sinx\right|dx}\\ 2I={\displaystyle \underset{-\frac{\pi }{4}}{\overset{\frac{\pi }{4}}{\int }}log\left|\left(cosx+sinx\right)\left(cosx-sinx\right)\right|dx}\\ 2I={\displaystyle \underset{-\frac{\pi }{4}}{\overset{\frac{\pi }{4}}{\int }}log\left|co{s}^{2}x-si{n}^{2}x\right|dx}\\ \therefore 2I={\displaystyle \underset{-\frac{\pi }{4}}{\overset{\frac{\pi }{4}}{\int }}logcos2xdx}\\ 2I=2{\displaystyle \underset{0}{\overset{\frac{\pi }{4}}{\int }}logcos2xdx}\left[?{\displaystyle \underset{-a}{\overset{a}{\int }}f\left(x\right)dx=2{\displaystyle \underset{0}{\overset{a}{\int }}f\left(x\right)dxiff\left(-x\right)=f\left(x\right)}}\right]\\ \therefore I=\pi {\displaystyle \underset{0}{\overset{\frac{\pi }{4}}{\int }}logcos2xdx}\\ Put2x=tdx=\frac{dt}{2}\\ Whenx=0\therefore t=0;whenx=\frac{\pi }{4}\therefore t=\frac{\pi }{2}\\ I=\frac{1}{2}{\displaystyle \underset{0}{\overset{\frac{\pi }{2}}{\int }}logcostdt}\text{\&thi}\end{array}$

$\begin{array}{l}Onadding\left(iii\right)and\left(iv\right),weget\\ 2I=\frac{1}{2}{\displaystyle \underset{0}{\overset{\frac{\pi }{2}}{\int }}\left(logcost+logsint\right)dt}\\ 2I=\frac{1}{2}{\displaystyle \underset{0}{\overset{\frac{\pi }{2}}{\int }}logsintcostdt}\\ 2I=\frac{1}{2}{\displaystyle \underset{0}{\overset{\frac{\pi }{2}}{\int }}\frac{log2sintcostdt}{2}}\\ 2I=\frac{1}{2}{\displaystyle \underset{0}{\overset{\frac{\pi }{2}}{\int }}\left(logsin2t-log2\right)dt}\\ 4I={\displaystyle \underset{0}{\overset{\frac{\pi }{2}}{\int }}logsin2tdt}-{\displaystyle \underset{0}{\overset{\frac{\pi }{2}}{\int }}log2dt}\\ Put2t=u\Rightarrow 2dt=du\Rightarrow dt=\frac{du}{2}\\ \therefore 4I=\frac{1}{2}{\displaystyle \underset{0}{\overset{\pi }{\int }}logsinudu}-{\displaystyle \underset{0}{\overset{\frac{\pi }{2}}{\int }}log2dt}\left[Changingthelimit\right]\\ 4I=\frac{1}{2}*2{\displaystyle \underset{0}{\overset{\frac{\pi }{2}}{\int }}logsinudu}-log2{\left[t\right]}_0^{\frac{\pi }{2}}\\ 4I={\displaystyle \underset{0}{\overset{\frac{\pi }{2}}{\int }}logsinudu}-log2.\frac{\pi }{2}\\ 4I=2I-\frac{\pi }{2}.log2\end{array}$

This is a Long answer type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l}LetI={\displaystyle \underset{0}{\overset{\pi }{\int }}xlogsinxdx}\dots \left(i\right)\\ ={\displaystyle \underset{0}{\overset{\pi }{\int }}\left(\pi -x\right)logsin\left(\pi -x\right)dx}\left[{\displaystyle \underset{0}{\overset{a}{\int }}f\left(x\right)dx={\displaystyle \underset{0}{\overset{a}{\int }}f\left(a-x\right)dx}}\right]\\ ={\displaystyle \underset{0}{\overset{\pi }{\int }}\left(\pi -x\right)logsinxdx}\dots \left(ii\right)\\ Adding\left(i\right)and\left(ii\right)\\ 2I={\displaystyle \underset{0}{\overset{\pi }{\int }}\left[\left(\pi -x\right)logsinx+xlogsinx\right]dx}\\ 2I={\displaystyle \underset{0}{\overset{\pi }{\int }}\pi logsinxdx}\\ 2I=2\pi {\displaystyle \underset{0}{\overset{\frac{\pi }{2}}{\int }}logsinxdx\left[?{\displaystyle \underset{0}{\overset{a}{\int }}f\left(x\right)dx=2{\displaystyle \underset{0}{\overset{a/2}{\int }}f\left(x\right)dx}}\right]}\\ \therefore I=\pi {\displaystyle \underset{0}{\overset{\frac{\pi }{2}}{\int }}logsinxdx}\dots \left(iii\right)\\ I=\pi {\displaystyle \underset{0}{\overset{\frac{\pi }{2}}{\int }}logsin\left(\frac{\pi }{2}-x\right)dx}\\ I=\pi {\displaystyle \underset{0}{\overset{\frac{\pi }{2}}{\int }}logcosxdx}\dots \left(iv\right)\\ Onadding\left(iii\right)and\left(iv\right),weget\\ 2I=\pi {\displaystyle \underset{0}{\overset{\frac{\pi }{2}}{\int }}\left(logsinx+logcosx\right)dx}\\ 2I=\pi {\displaystyle \underset{0}{\overset{\frac{\pi }{2}}{\int }}logsinxcosxdx}\\ 2I=\pi {\displaystyle \underset{0}{\overset{\frac{\pi }{2}}{\int }}\frac{log2sinxcosx}{2}dx}\\ 2I=\pi {\displaystyle \underset{0}{\overset{\frac{\pi }{2}}{\int }}logsin2xdx}-\pi {\displaystyle \underset{0}{\overset{\frac{\pi }{2}}{\int }}log2dx}\\ Put2x=t\Rightarrow 2dx=dt\Rightarrow dx=\frac{dt}{2}\\ 2I=\pi {\displaystyle \underset{0}{\overset{\pi }{\int }}logsintdt}-\pi .log2{\displaystyle \underset{0}{\overset{\frac{\pi }{2}}{\int }}1dx}\left[Changingthe\text{\hspace{0.17em}limit}\right]\\ 2I=I-\pi .log2{\left[x\right]}_0^{\frac{\pi }{2}}\text{\&}\end{array}$

This is a Long answer type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l}LetI={\displaystyle \underset{0}{\overset{1}{\int }}\underset{II}{xlog}}\underset{I}{\left|1+2x\right|}dx\\ ={\left[log\left|1+2x\right|.\left(\frac{x^2}{2}\right)\right]}_0^1-{\displaystyle \underset{0}{\overset{1}{\int }}\left(\frac{1.2}{1+2x}.\frac{x^2}{2}\right)}dx\\ =\frac{1}{2}{\left[x^{2}log\left(1+2x\right)\right]}_0^1-{\displaystyle \underset{0}{\overset{1}{\int }}\left(\frac{x^2}{1+2x}\right)}dx\\ =\frac{1}{2}\left[log3-0\right]-{\displaystyle \underset{0}{\overset{1}{\int }}\left(\frac{x}{2}-\frac{x/2}{1+2x}\right)}dx\\ =\frac{1}{2}log3-\frac{1}{2}{\displaystyle \underset{0}{\overset{1}{\int }}x}dx+\frac{1}{2}{\displaystyle \underset{0}{\overset{1}{\int }}\frac{x}{1+2x}}dx\\ =\frac{1}{2}log3-\frac{1}{2}{\left[\frac{x^2}{2}\right]}_0^1+\frac{1}{2}.\frac{1}{2}{\displaystyle \underset{0}{\overset{1}{\int }}\frac{\left(2x+1-1\right)}{1+2x}}dx\\ =\frac{1}{2}log3-\frac{1}{4}\left[1-0\right]+\frac{1}{4}{\displaystyle \underset{0}{\overset{1}{\int }}1dx}-\frac{1}{4}{\displaystyle \underset{0}{\overset{1}{\int }}\frac{1}{2x+1}dx}\\ =\frac{1}{2}log3-\frac{1}{4}+\frac{1}{4}{\left[x\right]}_0^1-\frac{1}{4}.\frac{1}{2}{\left[log\left|2x+1\right|\right]}_0^1\\ =\frac{1}{2}log3-\frac{1}{4}+\frac{1}{4}-\frac{1}{8}\left[log3-0\right]\\ =\frac{1}{2}log3-\frac{1}{8}log3=\frac{3}{8}log3\\ Hence,I=\frac{3}{8}log3.\end{array}$