Integrals

$ta{n}^{4}x=ta{n}^{2}xta{n}^{2}x$

$=(se{c}^{2}x-1)ta{n}^{2}x$

$=se{c}^{2}x.ta{n}^{2}x-ta{n}^{2}x$

$=se{c}^{2}x.ta{n}^{2}x-(se{c}^{2}x-1)$

I $\begin{array}{l}\begin{array}{}\end{array}=se{c}^{2}x.ta{n}^{2}x-se{c}^{2}x+1\end{array}$

$\begin{array}{l} I={\displaystyle \int ta{n}^4}xdx={\displaystyle \int se{c}^2}x.ta{n}^{2}xdx-{\displaystyle \int se{c}^2}xdx+{\displaystyle \int 1}.dx\end{array}$

$\begin{array}{l}={\displaystyle \int se{c}^2}x.ta{n}^{2}xdx-tan\pi +x+C-----\left(i\right)\end{array}$

$\begin{array}{l}LetI={\displaystyle \int se{c}^2}x.ta{n}^{2}xdx\end{array}$

$\begin{array}{l}Puttanx=t\end{array}$

$\begin{array}{l}se{c}^{2}xdx=dt\end{array}$

$\begin{array}{l}{I}_1={\displaystyle \int se{c}^2}x.ta{n}^{2}xdx\end{array}$

$\begin{array}{l}={\displaystyle \int t^2}dt\end{array}$

$=\frac{t^3}{3}=\frac{ta{n}^{3}x}{}$

$\frac{3}{}I={\displaystyle \int ta{n}^4}xdx$

$=\frac{1}{3}ta{n}^{3}x-tanx+x+\; C$

$\begin{array}{l}\begin{array}{l}ta{n}^{3}2xsec 2x= ta{n}^{2}2xtan 2xsec 2x\hfill \end{array}\end{array}$

$\begin{array}{l}\begin{array}{l} \{se{c}^2(2x)-1\} tan 2xsec 2x\hfill \end{array}\end{array}$

$\begin{array}{l}\begin{array}{l}= se{c}^{2}2xtan 2xsec 2x\hfill \end{array}\end{array}$

$\begin{array}{l}\begin{array}{l}-tan 2xsec 2x\hfill \end{array}\end{array}$

$\begin{array}{l}\begin{array}{l}I ={\displaystyle \int ta{n}^3}2x.sec 2x dx\hfill \end{array}\end{array}$

$\begin{array}{l}\begin{array}{l}={\displaystyle \int ta{n}^2}2xtan 2xsec 2x dx-{\displaystyle \int tan}2x.sec 2x dx\hfill \end{array}\\ ={\displaystyle \int ta{n}^2}2xtan2xsec2xdx-\frac{sec2x}{2}+\\ \begin{array}{l}Put sec 2x=t\hfill \end{array}\end{array}$

$\begin{array}{l}\begin{array}{l}\Rightarrow 2sec 2xtan 2x dx=dt\hfill \end{array}\end{array}$

$\begin{array}{l}\begin{array}{l}I={{\displaystyle \int tan}}^{3}2x.sec 2x dx\hfill \end{array}\\ =\frac{1}{2}{\displaystyle \int t^2}dt-\frac{sec2x}{2}+\; C\\ =\frac{t^3}{6}-\frac{sec2x}{2}+\; C\\ =\frac{{(sec2x)}^3}{6}-\frac{sec2x}{2}+\; C\end{array}$

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

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

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

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

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

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

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}$

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

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