Question

The value of $\underset{0}{\overset{\mathrm{\pi }}{\int }}\frac{x\tan x}{\sec x+\cos x}dx$ is
(a) $\frac{{\mathrm{\pi }}^2}{4}$

(b) $\frac{{\mathrm{\pi }}^2}{2}$

(c) $\frac{3{\mathrm{\pi }}^2}{2}$

(d) $\frac{{\mathrm{\pi }}^2}{3}$

Answer 1

$\left(\mathrm{a}\right)\frac{{\mathrm{\pi }}^2}{4}$π24

$$\begin{gathered} \mathrm{Let}I={\int }_0^{\mathrm{\pi }}\frac{x\tan x}{secx+\cos x}dx\left(i\right) \\ ={\int }_0^{\mathrm{\pi }}\frac{\left(\mathrm{\pi }-x\right)\tan \left(\mathrm{\pi }-x\right)}{sec\left(\mathrm{\pi }-x\right)+\cos \left(\mathrm{\pi }-x\right)}dx \\ ={\int }_0^{\mathrm{\pi }}\frac{\left(\mathrm{\pi }-x\right)\mathrm{tanx}}{secx+\cos x}dx\left(ii\right) \\ \mathrm{Adding}\left(i\right)\mathrm{and}\left(ii\right) \\ 2I={\int }_0^{\mathrm{\pi }}\frac{x\tan x}{secx+\cos x}+\frac{\left(\mathrm{\pi }-x\right)\mathrm{tanx}}{secx+\cos x}dx \\ ={\int }_0^{\mathrm{\pi }}\frac{\mathrm{\pi }\tan x}{secx+\cos x}dx \\ =\mathrm{\pi }{\int }_0^{\mathrm{\pi }}\frac{\mathrm{sinx}}{1+{\cos }^2\mathrm{x}}\mathrm{dx} \\ =-\mathrm{\pi }{\left[{\tan }^{-1}\mathrm{cosx}\right]}_0^{\mathrm{\pi }} \\ =-\mathrm{\pi }\left(-\frac{\mathrm{\pi }}{4}-\frac{\mathrm{\pi }}{4}\right)=\frac{{\mathrm{\pi }}^2}{2} \\ \mathrm{Hence}\mathrm{I}=\frac{{\mathrm{\pi }}^2}{4} \end{gathered}$$$$\begin{gathered} \mathrm{We}\mathrm{have}, \\ I={\int }_0^{\mathrm{\pi }}\frac{x\tan x}{\sec x+\cos x}\mathit{d}x\mathit{}.....\left(1\right) \\ ={\int }_0^{\mathrm{\pi }}\frac{\left(\mathrm{\pi }-x\right)\tan \left(\mathrm{\pi }-x\right)}{\sec \left(\mathrm{\pi }-x\right)+\cos \left(\mathrm{\pi }-x\right)}\mathit{d}x \\ ={\int }_0^{\mathrm{\pi }}\frac{\left(\mathrm{\pi }-x\right)\mathrm{tanx}}{\sec x+\cos x}dx.....\left(2\right) \\ \mathrm{Adding}\left(1\right)\mathrm{and}\left(2\right),\mathrm{we}\mathrm{get} \\ 2I={\int }_0^{\mathrm{\pi }}\left[\frac{x\tan x}{\sec x+\cos x}+\frac{\left(\mathrm{\pi }-x\right)\mathrm{tanx}}{\sec x+\cos x}\right]\mathit{d}x \\ \Rightarrow I=\frac{1}{2}{\int }_0^{\mathrm{\pi }}\frac{\mathrm{\pi }\tan x}{\sec x+\cos x}dx \\ =\frac{\mathrm{\pi }}{2}{\int }_0^{\mathrm{\pi }}\frac{\mathrm{sinx}}{1+{\cos }^2\mathrm{x}}\mathrm{dx} \\ \mathrm{Putting}\cos x=\mathrm{t} \\ \Rightarrow -\sin x\mathit{}dx=dt \\ \Rightarrow \sin x\mathit{}dx=-dt \\ \mathrm{When}x\to 0;t\to 1 \\ \mathrm{and}x\to \mathrm{\pi };t\to -1 \\ \Rightarrow I=\frac{\mathrm{\pi }}{2}{\int }_1^{-1}\frac{-dt}{1+{\mathrm{t}}^2} \\ =\frac{\mathrm{\pi }}{2}{\int }_{-1}^1\frac{dt}{1+{\mathrm{t}}^2} \\ =\frac{\mathrm{\pi }}{2}{\left[{\tan }^{-1}t\right]}_{-1}^1 \\ =\frac{\mathrm{\pi }}{2}\left[{\tan }^{-1}\left(1\right)-{\tan }^{-1}\left(-1\right)\right] \\ =\frac{\mathrm{\pi }}{2}\left[\frac{\mathrm{\pi }}{4}-\left(-\frac{\mathrm{\pi }}{4}\right)\right] \\ =\frac{\mathrm{\pi }}{2}\times \frac{\mathrm{\pi }}{2}=\frac{{\mathrm{\pi }}^2}{4} \\ \mathrm{Hence}\mathit{}I=\frac{{\mathrm{\pi }}^2}{4} \end{gathered}$$