# Methods of integration

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# Let ${\mathrm{x}}^{2}\ne \mathrm{n\pi }-1,\mathrm{n}\in \mathrm{N}$ . If $\int \mathrm{x}\sqrt{\frac{2\mathrm{sin}\left({\mathrm{x}}^{2}+1\right)-\mathrm{sin}2\left({\mathrm{x}}^{2}+1\right)}{2\mathrm{sin}\left({\mathrm{x}}^{2}+1\right)+\mathrm{sin}2\left({\mathrm{x}}^{2}+1\right)}}\mathrm{dx}=\mathrm{ln}\mathrm{f}\left(\mathrm{x}\right)+\mathrm{C}$ then ${\mathrm{sec}}^{-1}\mathrm{f}\left(1\right)$ equals

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Solution

## Let ${\mathrm{x}}^{2}+1=\mathrm{t}⇒\mathrm{xdx}=\frac{\mathrm{dt}}{2}$Given integral $=\frac{1}{2}\int \sqrt{\frac{2\mathrm{sin}\mathrm{t}-2\mathrm{sin}\mathrm{tcos}\mathrm{t}}{2\mathrm{sin}\mathrm{t}+2\mathrm{sin}\mathrm{tcos}\mathrm{t}}}\mathrm{dt}$$\begin{array}{l}=\frac{1}{2}\int \mathrm{tan}\frac{\mathrm{t}}{2}\mathrm{dt}=\mathrm{ln}\left|\mathrm{sec}\frac{\mathrm{t}}{2}\right|+\mathrm{C}\\ \therefore \mathrm{f}\left(\mathrm{x}\right)=\mathrm{sec}\left(\frac{{\mathrm{x}}^{2}+1}{2}\right)\\ {\mathrm{sec}}^{-1}\mathrm{f}\left(1\right)=1\end{array}$

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If $\int \frac{1-{x}^{9}}{x\left(1+{x}^{9}\right)}dx=A\mathrm{log}|x|+B\mathrm{log}\left|1+{x}^{9}\right|+C,$ then the value of $\frac{A}{B}$ is equal to