Ncert Solutions Maths class 12th

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$\begin{array}{l}Put7-4x=t\\ -4dx=dt\\ I=\int se{c}^2(7-4x)dx\\ =\frac{-1}{4}\int se{c}^{2}tdt=\frac{-1}{4}(tant)+C\end{array}$

$\begin{array}{l}Put2x-3=t\\ 2dx=dt\\ I=\int ta{n}^2(2x-3)dx={\displaystyle \int \left[se{c}^2(2x-3)-1\right]dx}\\ =\frac{1}{2}\int \left(se{c}^{2}t\right)dt-{\displaystyle \int 1}dx\\ =\frac{1}{2}tant-x+C=\frac{1}{2}tan(2x-3)-x+C\end{array}$

$\begin{array}{l}Put{e}^{2x}+e^{-2x}=t\\ \begin{array}{l}\Rightarrow (2e^{2x}-2{e^{-}}^{2x})dx=dt\hfill \end{array}\end{array}$

$\begin{array}{l}\begin{array}{ll}\Rightarrow 2(e^{2x}-{e^{-}}^{2x})dx=dt\hfill & \hfill \end{array}\\ I=\int \frac{e^{2x}-e^{-2x}}{e^{2x}+e^{-2x}}dx=\int \frac{dt}{2t}=\frac{1}{2}\int \frac{1}{t}dt\\ =\frac{1}{2}log\left|t\right|+=\frac{1}{2}log\left|e^{2x}+e^{-2x}\right|+C\end{array}$

Dividing both numerator and denominator by e^x, we get

$\begin{array}{l}\frac{\frac{e^{2x}-1}{e^x}}{\frac{e^{2x}+1}{e^x}}=\frac{e^x-e^{-x}}{e^x+e^{-x}}\\ Put{e}^x+e^{-x}=t\\ \Rightarrow \left(e^x-e^{-x}\right)dx=dt\\ I=\int \frac{e^{2x}-1}{e^{2x}+1}dx=\int \frac{e^x-e^{-x}}{e^x+e^{-x}}dx\\ I=\int \frac{dt}{t}=log\left|t\right|+c\\ =log\left|e^x+e^{-x}\right|+c\end{array}$

$\begin{array}{l}Putta{n}^{-1}x=t\\ \frac{1dx}{1+x^2}=dt\\ I=\int \frac{e^{ta{n}^{-1}x}}{1+x^2}dx=\int e^{t}dt=e^t+c\\ =e^{ta{n}^{-1}x}+c\end{array}$

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

$\begin{array}{l}Put2x+3=t\\ 2dx=dt\\ I=\int e^{2x+3}dx\\ =\frac{1}{2}\int e^t·dt\\ =\frac{1}{2}\left(e^t\right)+C\\ =\frac{1}{2}{e}^{(2x+3)}+C\end{array}$

$\begin{array}{l}Put9-4x^2=t\\ -8xdx=dt\\ xdx=\frac{-1}{8}dt\\ -\frac{1}{8}{\displaystyle \int \frac{dt}{t}}=-\frac{1}{8}log\left|t\right|+c=-\frac{1}{8}log\left|9-4x^2\right|+c\end{array}$

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