Ncert Solutions Maths class 12th

$\begin{array}{l}I={\displaystyle \int si{n}^3\left(2x+1\right)dx}\\ ={\displaystyle \int si{n}^2\left(2x+1\right).sin}\left(2x+1\right)dx\\ ={\displaystyle \int \left\{1-co{s}^2\left(2x+1\right)\right\}}sin\left(2x+1\right)dx\\ Puttingcos\left(2x+1\right)=t\\ \Rightarrow -2sin\left(2x+1\right)dx=dt\\ \Rightarrow sin\left(2x+1\right)dx=\frac{-dt}{2}\\ I=\frac{-1}{2}{\displaystyle \int \left(1-t^2\right)}dt\\ =\frac{-1}{2}\left\{t-\frac{t^3}{3}\right\}+\; C\\ =\frac{-1}{2}\left\{cos(2x+1)-\frac{co{s}^3(2x+1){}^3}{3}\right\}+\; C\\ =\frac{-cos(2x+1)}{2}+\frac{co{s}^3(2x+1){}^6}{6}+\; C\\ =\frac{-1}{2}cos(2x+1)+\frac{1}{6}co{s}^3(2x+1)+\; C\end{array}$

$\begin{array}{l}Here,cosAcosB=\frac{1}{2}\{cos(A+B)+cos(A-B)\}\\ I={\displaystyle \int cos2x\left(cos4xcos6x\right)dx}\\ ={\displaystyle \int cos}2x\left[\frac{1}{2}cos(4x+6x)+cos(4x-6x)\right]dx\\ ={\displaystyle \int cos2x\left[\frac{1}{2}(cos10x+cos(-2x))\right]dx}\\ =\frac{1}{2}{\displaystyle \int cos}2xcos10x+cos2xcos(-2x)dx\\ =\frac{1}{2}{\displaystyle \int cos}2xcos10x+cos2xdx\left[\therefore cos\left(-x\right)=cosx\right]\\ =\frac{1}{2}\left[\frac{1}{2}\{cos2x+10x+cos2x-10\}+\left\{\frac{1+cos4x}{2}\right\}\right]dx\\ =\frac{1}{4}{\displaystyle \int cos}12x+cos8x+1+cos4x\cdot dx\\ =\frac{1}{4}[\frac{sin12x}{12}+\frac{sin8x}{8}+x+\frac{cos4x}{4x}]+\; C\end{array}$

$\begin{array}{l}Here,\\ sinAcosB=\frac{1}{2}\left\{sin(A+B)+sin(A-B)\right\}\\ sin3xcos4x=\frac{1}{2}(sin(3x+4x)+sin(3x-4x))\\ Then,\\ {\displaystyle \int sin}3xcos4xdx={\displaystyle \int \frac{1}{2}}[sin(3x+4x)+sin\left(3x-4x\right)dx\\ =\frac{1}{2}{\displaystyle \int sin}(3x+4x)+sin\left(3x-4x\right)dx\\ =\frac{1}{2}{\displaystyle \int sin}7x+sin(-x)dx\\ =\frac{1}{2}\left[-\frac{cos7x}{7}+cosx\right]+c\\ =\frac{-cos7x}{14}+\frac{cosx}{2}+c\end{array}$

$\begin{array}{l}Here,\\ si{n}^2\left(2x+5\right)=\frac{1-cos2(2x+5)}{2}\\ =\frac{1-cos(4x+10)}{2}\\ Then,\\ ={\displaystyle \int si{n}^2}(2x+5)dx={\displaystyle \int \frac{1-cos(4x+10)}{2}}dx\\ =\frac{1}{2}{\displaystyle \int 1}\cdot dx-\frac{1}{2}{\displaystyle \int cos}(4x+10)dx\\ =\frac{1}{2}\cdot x-\frac{1}{2}\frac{sin(4x+10)}{4}+\; C\\ =\frac{x}{2}-\frac{1}{8}sin(4x+10)+\; C\end{array}$

$\begin{array}{l}I=\int \frac{dx}{si{n}^{2}xco{s}^{2}x}\\ =\int \frac{1}{si{n}^{2}xco{s}^{2}x}dx\\ =\int \frac{si{n}^{2}x+co{s}^{2}x}{si{n}^{2}xco{s}^{2}x}dx\\ =\int \frac{si{n}^{2}x}{si{n}^{2}xco{s}^{2}x}dx+\int \frac{co{s}^{2}x}{si{n}^{2}xco{s}^{2}x}dx\\ =\int se{c}^{2}xdx+\int cose{c}^{2}x-dx]\\ =tanx-cotx+c\end{array}$

Therefore, the correct answer is B.

$\begin{array}{l}Put{x}^{10}+10^x=t\\ \Rightarrow \left(10x^9+10^{x}lo{g}^{e}10\right)dx=dt\\ I=\int \frac{10x^9+10^{x}lo{g}_{e}10}{x^{10}+10^x}dx\\ =\int \frac{dt}{t}=logt+C\\ =log\left(10^x+x^{10}\right)+c\end{array}$

Therefore, the correct answer is (D)

$\begin{array}{l}Put{x}^4=t\\ 4x^{3}dx=dt\\ I=\int \frac{x^{3}sin\left(ta{n}^{-1}{x}^4\right)}{1+x^8}dx\\ =\frac{1}{4}\int \frac{sin\left(ta{n}^{-1}t\right)}{1+t^2}\_\_\_\_\_(1)\\ Putta{n}^{-1}t=u\\ \Rightarrow \frac{1}{1+t^2}dt=du\end{array}$

From (1), we get

$\begin{array}{l}I=\int \frac{x^{3}sin\left(ta{n}^{-1}{x}^4\right)}{1+x^8}dx\\ =\frac{1}{4}{\displaystyle \int sin}udu\\ =\frac{1}{4}(-cosu)+C\\ =-\frac{1}{4}cos\left(ta{n}^{-1}t\right)+C\\ =-\frac{1}{4}cos\left(ta{n}^{-1}{x}^4\right)+C\end{array}$

$\begin{array}{l}I=\frac{(x+1)}{x}{(x+logx)}^2\\ =\left(1+\frac{1}{x}\right){(x+logx)}^2\\ Putx+logx=t\\ \Rightarrow 1+\frac{1}{x}dx=dt\\ I=\int \left(1+\frac{1}{x}\right){(x+logx)}^{2}dx\\ =\int t^{2}dt=\frac{t^3}{3}+C\\ =\frac{{(x+logx)}^3}{3}+C\end{array}$

$\begin{array}{l}Put\left(1+logx\right)=t\\ \Rightarrow \frac{1}{x}dx=dt\\ I=\int \frac{{(1+logx)}^2}{x}dx\\ =\int t^{2}dt\\ =\frac{t^3}{3}+C\frac{{(1+logx)}^3}{3}+C\end{array}$

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