Question

\[\int\limits_0^\pi \frac{x}{a^2 - \cos^2 x} dx, a > 1\]

Answer 1

\[Let I = \int_0^\pi \frac{x}{a^2 - \cos^2 x} d x ..............(1)\]

\[ = \int_0^\pi \frac{\pi - x}{a^2 - \cos^2 \left( \pi - x \right)} d x \]

\[ = \int_0^\pi \frac{\pi - x}{a^2 - \cos^2 x} d x ...............(2)\]

Adding (1) and (2)

\[2I = \int_0^\pi \frac{\pi}{a^2 - \cos^2 x} d x \]

\[ = \frac{\pi}{2a} \int_0^\pi \left[ \frac{1}{a - cosx} + \frac{1}{a + cosx} \right] dx\]

\[ = \frac{\pi}{2a} \int_0^\pi \left[ \frac{\sec^2 \frac{x}{2}}{\left( a - 1 \right) + \left( a + 1 \right) \tan^2 \frac{x}{2}} + \frac{\sec^2 \frac{x}{2}}{\left( a + 1 \right) + \left( a - 1 \right) \tan^2 \frac{x}{2}} \right]dx\]

\[Let, \tan\frac{x}{2} = t, then \frac{1}{2} \sec^2 \frac{x}{2} dx = dt\]

\[2I = \frac{\pi}{a} \int_0^\infty \left[ \frac{1}{\left( a - 1 \right) + \left( a + 1 \right) t^2} + \frac{1}{\left( a + 1 \right) + \left( a - 1 \right) t^2} \right] dt\]

\[ = \frac{\pi}{a\sqrt{\left( a^2 - 1 \right)}} \left[ \tan^{- 1} \sqrt{\frac{a + 1}{a - 1}}t + \tan^{- 1} \sqrt{\frac{a - 1}{a + 1}}t \right]_0^\infty \]

\[ = \frac{\pi}{a\sqrt{\left( a^2 - 1 \right)}}\left[ \frac{\pi}{2} + \frac{\pi}{2} \right]\]

\[ = \frac{\pi^2}{a\sqrt{\left( a^2 - 1 \right)}}\]

\[ \therefore I = \frac{\pi^2}{2a\sqrt{\left( a^2 - 1 \right)}}\]