Ncert Solutions Maths class 12th

$\begin{array}{l}\frac{cosx-sinx}{1+sin2x}=\frac{cosx-sinx}{\left(si{n}^{2}x+co{s}^{2}x\right)+2sinxcosx}\\ =\frac{cosx-sinx}{{(sinx+cosx)}^2}\\ \left[?si{n}^2+co{s}^2=1\&sin2x=2sinxcosx\right]\\ I={\displaystyle \int \frac{cosx-sinx}{1+sin2x}}dx={\displaystyle \int \frac{cosx-sinx}{{(sinx+cosx)}^2}dx}\\ \begin{array}{l}Put sinx+ cosx=t\hfill \end{array}\end{array}$

$\begin{array}{l}\begin{array}{l}(cosx-sinx)dx=dt\hfill \end{array}\\ ={\displaystyle \int \frac{dt}{t^2}}={\displaystyle \int t^{-2}}dt\\ =-t^{-1}+=-\frac{1}{t}+\; c\\ =\frac{1}{-sinx+cosx}+\; c\end{array}$

$\begin{array}{l}=4cos\left(\frac{x+\propto }{2}\right)cos\left(\frac{x-\propto }{2}\right)\\ =2[cos\left(\frac{x+\propto }{2}+\frac{x-\propto }{2}\right)+cos\left(\frac{x-\propto }{2}-\frac{x-\propto }{2}\right)\\ \begin{array}{ll}= 2[cos\left(x\right) + cos\propto ]\hfill & \begin{array}{l}=2cosx+2cos\propto \\ ={\displaystyle \int \frac{cos2x-cos2\propto }{cosx-cos\propto }}dx\\ \begin{array}{l}={\displaystyle \int (2cosx+2cos\propto )}dx\hfill \end{array}\end{array}\hfill \end{array}\end{array}$

$\begin{array}{l}\begin{array}{l}\begin{array}{l}\begin{array}{l}= 2[sinx+xcos\propto ] + C\hfill \end{array}\end{array}\hfill \end{array}\end{array}$

$\frac{2^{}x}{1+cosx}=\frac{{\left(2sin\frac{x}{2}\cdot cos\frac{x}{2}\right)}^2}{2co{s}^2\frac{x}{2}}$

$\begin{array}{l}=\frac{4si{n}^2\frac{x}{2}co{s}^2\frac{x}{2}}{2co{s}^2\frac{x}{2}}=\frac{2si{n}^2\frac{x}{2}}{2}\\ =1-cosx\\ ={\displaystyle \int \frac{si{n}^{2}x}{1+cosx}}dx\\ ={\displaystyle \int (1-cosx)}dx\\ =x-sinx+C\end{array}$

Kindly go through the solution

$\begin{array}{l}si{n}^{4}x=si{n}^{2}x.si{n}^{2}x\\ =\left(\frac{1-cos2x}{2}\right)\left(\frac{1-cos2x}{2}\right)\\ =\frac{1}{4}{(1-cos2x)}^2=\frac{1}{4}\left[1+co{s}^{2}2x-2cos2x\right]\\ =\frac{1}{4}\left[1+\left(\frac{1+cos4x}{2}\right)-2cos2x\right]\\ =\frac{1}{4}\left[1+\frac{1}{2}+\frac{1}{2}cos4x-2cos2x\right]\\ =\frac{1}{4}\left[\frac{3}{2}+\frac{1}{2}cos4x-2cos2x\right]\end{array}$

$\begin{array}{l}={\displaystyle \int si{n}^4}xdx=\frac{1}{4}{\displaystyle \int [\frac{3}{2}}+\frac{1}{2}cos4x-2cos2x]dx\\ =\frac{1}{4}\left[\frac{3}{2}+\frac{1}{2}\left(\frac{sin4x}{4}\right)-\frac{2sin2x}{2}\right]+\; C\\ =\frac{1}{8}\left[3x+\frac{si{n}^{4}x}{4}-2sin2x\right]+\; C\\ =\frac{3x}{8}-\frac{1}{4}sin2x+\frac{1}{32}sin4x+\; C\end{array}$

$\begin{array}{l}\frac{cosx}{1+cosx}=\frac{co{s}^2\frac{x}{2}-si{n}^2\frac{x}{2}}{2co{s}^2\frac{x}{2}}\\ =\frac{1}{2}\left[1-ta{n}^2\frac{x}{2}\right]\\ ={\displaystyle \int \frac{cosx}{1+cosx}}dx=\frac{1}{2}{\displaystyle \int \left(1-ta{n}^2\frac{x}{2}\right)}dx\\ =\frac{1}{2}\left(1-se{c}^2\frac{x}{2}+1\right)dx\\ =\frac{1}{2}\left(2-se{c}^2\frac{x}{2}\right)dx=\frac{1}{2}\left[2x-\frac{tan\frac{x}{2}}{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\right]+\; C\\ =x-tan\frac{x}{2}+\; C\end{array}$

$\begin{array}{l}\frac{1-cosx}{1+cosx}=\frac{2si{n}^2\frac{x}{2}}{2co{s}^2\frac{x}{2}}\\ =ta{n}^2\frac{x}{2}=se{c}^2\frac{x}{2}-1\\ ={\displaystyle \int \frac{1-cosx}{1+cosx}}dx={\displaystyle \int \left(se{c}^2\frac{x}{2}-1\right)}dx\\ =\left[\frac{tan\frac{x}{2}}{\frac{1}{2}}\right]+\; C\\ =2tan\frac{x}{2}+\; C\end{array}$

$\begin{array}{l}sinAsinB=\frac{-1}{2}\left\{cos(A+B)-cos(A-B)\right\}\\ ={\displaystyle \int sin}4xsin8xdx={\displaystyle \int \frac{1}{2}}\{cos(4x-8x)-cos(4x+8x)\}dx\\ =\frac{1}{2}{\displaystyle \int cos}(-4x)-cos12xdx\\ =\frac{1}{2}{\displaystyle \int (cos4x-cos12x)}dx\\ =\frac{1}{2}\left[\frac{sin4x}{4}-\frac{sin12x}{12}\right]+\; C\end{array}$

$\begin{array}{l}Here,\\ sinAsinB=\frac{-1}{2}\{cos(A+B)-cos(A-B)\}\\ I={\displaystyle \int sinx}sin2xsin3x\\ ={\displaystyle \int \left[sinx\cdot \frac{1}{2}\{cos(2x-3x)-cos(2x+3x)\}\right]}dx\\ ={\displaystyle \int \{[sin}x\cdot \frac{1}{2}\{cos(-x)-cos5x\}\}dx\\ =\frac{1}{2}{\displaystyle \int sin}xcosx-sinxcos5xdx\\ Now,sin2x=2sinxcosx,\\ =\frac{1}{2}{\displaystyle \int \frac{sin2x}{2}}dx-\frac{1}{2}{\displaystyle \int sin}xcos5xdx\\ =\frac{1}{4}\left[\frac{-cos2x}{2}\right]-\frac{1}{2}{\displaystyle \int \frac{1}{2}}sin(x+5x)+sin(x-5x)dx\\ =\frac{-cos2x}{8}-\frac{1}{4}{\displaystyle \int sin}6x+sin(-4x)dx\\ =\frac{-cos2x}{8}-\frac{1}{4}\left[\frac{-cos6x}{6}+\frac{cos4x}{4}\right]+\; C\\ =\frac{-cos2x}{8}-\frac{1}{8}\left[\frac{-cos6x}{3}+\frac{cos4x}{2}\right]+\; C\\ =\frac{-cos2x}{8}+\frac{cos6x}{24}-\frac{cos4x}{16}+\; C\\ =\frac{1}{4}\left[\frac{1}{6}cos6x-\frac{cos4x}{4}-\frac{cos2x}{2}\right]+\; C.\end{array}$

$\begin{array}{l}I={\displaystyle \int si{n}^{3}x}co{s}^{3}xdx\\ ={\displaystyle \int co{s}^3}xsi{n}^{2}xsinx.dx\\ ={\displaystyle \int co{s}^3}x(1-co{s}^{2}x).sinxdx\\ Putcosx=t\\ \Rightarrow -sin.x.dx=dt\\ I=-{\displaystyle \int t^3}(1-t^2)dt\end{array}$

$\begin{array}{l}=-{\displaystyle \int \left(t^3-t^5\right)}dt=-\left\{\frac{t^4}{4}-\frac{t^6}{6}\right\}+\; C\\ =-\left\{\frac{co{s}^{4}x}{4}-\frac{co{s}^{6}x}{6}\right\}+\; C\\ =\frac{-co{s}^{4}x}{4}+\frac{co{s}^{6}x}{6}+\; C\\ =\frac{1}{6}co{s}^{6}x-\frac{1}{4}co{s}^{4}x+\; C\end{array}$