Class 12th

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.

Yes, Chouksey Engineering College accept Class 12th marks for admission. The basic eligibility for course admission is to pass Class 12. Chouksey Engineering College offers BTech, MTech, and MBA under two streams, i.e., Management and Engineering. Apart from the UG and PG courses, the college also offers PhD courses. 

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:

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

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

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}$