Ncert Solutions Maths class 12th

Kindly go through the solution

Kindly go through the solution

$\begin{array}{l}\begin{array}{l}Let{x}^3=t\hfill \end{array}\end{array}$

$\begin{array}{l}\begin{array}{l}3x^{2}dx=dt\hfill \end{array}\\ I=\int \frac{3x^2}{x^6+1}dx=\int \frac{dt}{t^2+1}\\ \begin{array}{ll}= ta{n}^{–1}t+C\hfill & \hfill \end{array}\end{array}$

$\begin{array}{l}\begin{array}{l}= ta{n}^{–1}\left(x^3\right) +C\hfill \end{array}\end{array}$

$\begin{array}{l}\begin{array}{l}Let{e}^{2}x=t\hfill \end{array}\end{array}$

$\begin{array}{l}\begin{array}{l}{e}^{2}x+ e^{x}1dx=dt\hfill \end{array}\end{array}$

$\begin{array}{l}\begin{array}{l}{e}^x\left(x+1\right)dx=dt\hfill \end{array}\\ ={\displaystyle \int \frac{e^x(1+x)}{co{s}^2\left(e^{2}x\right)}}dx={\displaystyle \int \frac{dt}{co{s}^{2}t}}\\ \begin{array}{l}={\displaystyle \int se{c}^2}t dt = tan \left(e^x,x\right) + C\hfill \end{array}\end{array}$

$\begin{array}{l}\begin{array}{l}\therefore The correct answer is \left(B\right).\hfill \end{array}\end{array}$

$\begin{array}{l}={\displaystyle \int \frac{si{n}^{2}x-co{s}^{2}x}{si{n}^{2}co{s}^{2}x}}dx\\ \begin{array}{l}={\displaystyle \int (se{c}^{2}x-cose{c}^{2}x)}dx\hfill \end{array}\end{array}$

$\begin{array}{l}\begin{array}{l}= tanx+ cotx+ C.\hfill \end{array}\end{array}$

$\begin{array}{l}\begin{array}{l}Therefore, the correct answer is \left(A\right).\hfill \end{array}\end{array}$

$\begin{array}{l}\frac{1}{cos(x-a)cos(x-b)}=\frac{1}{sin(a-b)}*\left[\frac{sin(a-b)}{cos(x-a)cos(x-b)}\right]\\ =\frac{1}{sin(a-b)}\left[\frac{sin\{(x-b)-(x-a)\}}{cos(x-a)cos(x-b)}\right]\end{array}$

$\begin{array}{l}=\frac{1}{sin(a-b)}[sin(x-b)cos(x-a)-cos(x-b)\cdot sin(x-a)cos(x-a)cos(x-b)\\ =\frac{1}{sin(a-b)}[tan(x-b)-tan(x-a)]\\ =\frac{1}{sin(a-b)}{\displaystyle \int tan}(x-b)-tan\left(x-a\right)]dx\\ =\frac{1}{sin(a-b)}[-log|cos(x-b)|+log|cos(x-a)?]\\ =\frac{1}{sin(a-b)}\left[log\left|\frac{cos(x-a)}{cos(x-b)}\right|\right]+\; C\end{array}$

$\begin{array}{l}I={\displaystyle \int si{n^{-}}^1}\left(cosx\right)dx\\ ={\displaystyle \int si{n}^{-1}}\left(sin\left\{\frac{\pi }{2}-x\right\}\right)dx\\ ={\displaystyle \int \{\frac{\pi }{2}-x\}dx}\\ =\frac{\pi }{2}x-\frac{x^2}{2}+\; C\end{array}$

$\begin{array}{l}=\frac{cos2x}{co{s}^{2}x+si{n}^{2}x+2sinxcosx}\\ =\frac{cos2x}{1+sin2x}\end{array}$

$\begin{array}{l}={\displaystyle \int \frac{cos2x}{{(cosx+sinx)}^2}}dx\\ ={\displaystyle \int \frac{cos2x}{1+sin2x}}dx\\ \begin{array}{l}Put 1 + sin 2x=t\hfill \end{array}\end{array}$

$\begin{array}{l}\begin{array}{l}2 cos 2x dx=dt\hfill \end{array}\\ ={\displaystyle \int \frac{cos2x}{{(cosx+sinx)}^2}}dx\\ =\frac{1}{2}{\displaystyle \int \frac{1}{t}}dt=\frac{1}{2}log\left|t\right|+\; C\\ =\frac{1}{2}log\left|1+sin2x\right|+\; C\\ -\frac{1}{2}log\left|{(cosx+sinx)}^2\right|+\; C\\ =log\left|cosx+sinx\right|+C\end{array}$

$\begin{array}{l}\frac{si{n}^{2}x+co{s}^{2}x}{sinxco{s}^{3}x}=\frac{sinx}{co{s}^{3}x}+\frac{1}{sinxcosx}\\ =tanxse{c}^{2}x+\frac{co{s}^{2}x}{\left(\frac{sinxcosx}{co{s}^{2}x}\right)}\\ =tanxse{c}^{2}x+\frac{se{c}^{2}x}{tanx}\\ I={\displaystyle \int \frac{1}{sinxco{s}^{3}x}}dx={\displaystyle \int tan}xse{c}^{2}xdx+{\displaystyle \int \frac{se{c}^{2}x}{tanx}}dx\\ \begin{array}{l}Put tanx=t\hfill \end{array}\end{array}$

$\begin{array}{l}\begin{array}{l}Se{c}^{2}x dx=dt\hfill \end{array}\\ ={\displaystyle \int \frac{1}{sinxco{s}^{3}x}}dx={\displaystyle \int tanxse{c}^{2}xdx+{\displaystyle \int \frac{se{c}^{2}x}{tanx}}dx}\\ ={\displaystyle \int t}dt+{\displaystyle \int \frac{1dt}{t}}\\ =\frac{t^2}{2}+log\left|t\right|+\; C\\ =\frac{1}{2}ta{n}^{2}x+log\left|tanx\right|+\; C\end{array}$

$\begin{array}{l}=\frac{cos2x+(1-cos2x)}{co{s}^{2}x}\\ =\frac{1}{co{s}^{2}x}=se{c}^{2}x\\ ={\displaystyle \int \frac{cos2x+2si{n}^{2}x}{co{s}^{2}x}}dx\\ \begin{array}{l}={\displaystyle \int se{c}^2}xd x\hfill \end{array}\end{array}$

$\begin{array}{l}\begin{array}{l}= tanx+ C\hfill \end{array}\end{array}$