Question

The value of \[\int\limits_0^\pi \frac{x \tan x}{\sec x + \cos x} dx\] is __________ .

Options

• \[\frac{\pi^2}{4}\]
• \[\frac{\pi^2}{2}\]
• \[\frac{3 \pi^2}{2}\]

• \[\frac{\pi^2}{3}\]

Answer 1

\[ \frac{\pi^2}{4}\]

 

\[\text{We have}, \]

\[ I = \int_0^\pi \frac{x \tan x}{\sec x + \cos x} d x ..................(1)\]
\[ = \int_0^\pi \frac{\left( \pi - x \right)\tan\left( \pi - x \right)}{\sec\left( \pi - x \right) + \cos\left( \pi - x \right)} d x\]
\[ = \int_0^\pi \frac{\left( \pi - x \right)tanx}{\sec x + \cos x} dx .......................(2)\]
Adding (1) and (2), we get
\[2I = \int_0^\pi \left[ \frac{x\tan x}{\sec x + \cos x} + \frac{\left( \pi - x \right)tan x}{\sec x + \cos x} \right] d x\]
\[ \Rightarrow I = \frac{1}{2} \int_0^\pi \frac{\pi \tan x}{\sec x + \cos x}dx\]
\[ = \frac{\pi}{2} \int_0^\pi \frac{sin x}{1 + \cos^2 x} dx\]
\[\text{Putting} \cos x = t\]
\[ \Rightarrow - \sin x dx = dt\]
\[ \Rightarrow \sin x dx = - dt\]
\[When\ x \to 0; t \to 1\]
\[and\ x \to \pi; t \to - 1\]
\[ \Rightarrow I = \frac{\pi}{2} \int_1^{- 1} \frac{- dt}{1 + t^2}\]
\[ = \frac{\pi}{2} \int_{- 1}^1 \frac{dt}{1 + t^2}\]
\[ = \frac{\pi}{2} \left[ \tan^{- 1} t \right]_{- 1}^1 \]
\[ = \frac{\pi}{2}\left[ \tan^{- 1} \left( 1 \right) - \tan^{- 1} \left( - 1 \right) \right]\]
\[ = \frac{\pi}{2}\left[ \frac{\pi}{4} - \left( - \frac{\pi}{4} \right) \right]\]
\[ = \frac{\pi}{2} \times \frac{\pi}{2} = \frac{\pi^2}{4}\]
\[Hence\ I = \frac{\pi^2}{4}\]