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प्रश्न
\[\int\limits_0^\pi \frac{x \tan x}{\sec x + \tan x} dx\]
उत्तर
\[Let I = \int_0^\pi \frac{x \tan x}{sec x + \tan x} d x ...........(1)\]
\[ = \int_0^\pi \frac{\left( \pi - x \right) \tan\left( \pi - x \right)}{sec\left( \pi - x \right) + \tan\left( \pi - x \right)} d x\]
\[ = \int_0^\pi \frac{\left( \pi - x \right) \tan x}{\sec x + \tan x} d x ................(2)\]
Adding (1) and (2) we get
\[2I = \int_0^\pi \frac{\pi \tan x}{\sec x + \tan x} d x\]
\[ = \pi \int_0^\pi \frac{sinx}{1 + sin x}dx\]
\[ = \pi \int_0^\pi \frac{1 + sin x - 1}{1 + sin x}dx\]
\[ = \pi \int_0^\pi \left[ 1 - \frac{1}{1 + sinx} \right]dx\]
\[ = \pi \left[ x \right]_0^\pi - \pi \int_0^\pi \frac{1}{1 + \frac{2\tan\frac{x}{2}}{1 + \tan^2 \frac{x}{2}}}dx\]
\[ = \pi^2 - \pi \int_0^\pi \frac{\sec^2 \frac{x}{2}}{1 + \tan^2 \frac{x}{2} + 2\tan\frac{x}{2}}dx\]
\[ = \pi^2 - \pi \int_0^\pi \frac{\sec^2 \frac{x}{2}}{\left( 1 + \tan\frac{x}{2} \right)^2}dx\]
\[ = \pi^2 + \pi \left[ \frac{2}{1 + \tan\frac{x}{2}} \right]_0^\pi \]
\[ = \pi^2 + \pi\left( 0 - 2 \right)\]
\[ = \pi^2 - 2\pi\]
\[ = \pi\left( \pi - 2 \right)\]
\[\text{Hence }I = \frac{\pi}{2}\left( \pi - 2 \right)\]
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