Class 12th

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

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

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

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

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

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

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

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

This is a Multiple Choice Questions as classified in NCERT Exemplar

Answer- (a)

Explanation- when a positive point charge is brought near an isolated conducting sphere without touching the sphere, then the free electron in the sphere are attracted towards the positive charge. This leaves an excess positive charge on right side of sphere . And also by induction the negative charge is setup across left side. So field lines first goes from positive charge (+q) to negative charge of sphere then from positive charge of sphere to negative charge anywhere in the universe.

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

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

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

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

This is a Multiple Choice Questions as classified in NCERT Exemplar

Answer.   (a)

Explanation-The force between (q1, q2) and (q1, q3) is must be attractive to net forces act in positive direction of x . so q1 must be negative. But if we place a positive charge Q in positive direction of x then it must be attractive towards along x-axis. So it will be attractive towards positive x axis. 

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

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

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}$