Exemplar Solution

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

Sol:

$\begin{array}{l}LetI={\displaystyle \int \frac{dx}{1+cosx}}\\ ={\displaystyle \int \frac{dx}{2co{s}^{2}x/2}\left[?1+cosx=2co{s}^{2}x/2\right]}\\ =\frac{1}{2}{\displaystyle \int se{c}^2\frac{x}{2}dx}=\frac{1}{2}.2tan\frac{x}{2}+C=tan\frac{x}{2}+C\\ Hence,therequiredsolutionistan\frac{x}{2}+C.\end{array}$

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

Sol:

$\begin{array}{l}LetI={\displaystyle \int \frac{1+cosx}{x+sinx}}dx\\ Putx+sinx=t\Rightarrow \left(1+cosx\right)dx=dt\\ \therefore I={\displaystyle \int \frac{dt}{t}}=log\left|t\right|=log\left|x+sinx\right|+C\\ Hence,therequiredsolutionislog\left|x+sinx\right|+C.\end{array}$

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

Sol:

$\begin{array}{l}LetI={\displaystyle \int \frac{e^{6logx}-e^{5logx}}{e^{4logx}-e^{3logx}}}dx\\ \therefore I={\displaystyle \int \frac{e^{log{x}^6}-e^{log{x}^5}}{e^{log{x}^4}-e^{log{x}^3}}}dx\\ ={\displaystyle \int \frac{x^6-x^5}{x^4-x^3}dx}={\displaystyle \int \frac{x^2\left(x^4-x^3\right)}{x^4-x^3}dx}={\displaystyle \int x^{2}dx}\\ =\frac{1}{3}{x}^3+C\\ Hence,therequiredsolutionis\frac{1}{3}{x}^3+C.\end{array}$

This is a long answer type question as classified in NCERT Exemplar

Both prontosil and Salvarsan and antibacterial drugs (antimicrobials) discovered by Paul Ehrlich. Salvarsan is also known as arsphenamine. It is an organoarsenic molecule and has -As = As- double bond.

Salvarsan and prontosil show similarity in their structure. Both of these drugs are antimicrobials. Salvarsan contains -As = As- linkage whereas prontosil has -N = N- linkage.

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

Sol:

$\begin{array}{l}LetI={\displaystyle \int \frac{\left(x^2+2\right)}{x+1}}dx\\ \therefore I={\displaystyle \int \left[\left(x-1\right)+\frac{3}{x+1}\right]}dx\\ ={\displaystyle \int \left(x-1\right)dx}+3{\displaystyle \int \frac{1}{x+1}dx}\\ =\frac{x^2}{2}-x+3log\left|x+1\right|+C\\ Hence,therequiredsolutionis\frac{x^2}{2}-x+3log\left|x+1\right|+C.\end{array}$

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

Sol:

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

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:

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

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