Question

\[\int\limits_0^\pi \frac{x}{a^2 \cos^2 x + b^2 \sin^2 x} dx\]

Answer 1

We have,

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

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

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

Adding (1) and (2)

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

\[ = \pi \int_0^\pi \frac{1}{a^2 \cos^2 x + b^2 \sin^2 x} d x\]

\[ = \pi \int_0^\pi \frac{\sec^2 x}{a^2 + b^2 \tan^2 x}dx ...............\left(\text{Dividing numerator and denominator by }\cos^2 x \right)\]

\[ = 2\pi \int_0^\frac{\pi}{2} \frac{\sec^2 x}{a^2 + b^2 \tan^2 x}dx ..............\left[\text{Using }\int_0^{2a} f\left( x \right)dx = \int_0^a f\left( x \right)dx + \int_0^a f\left( 2a - x \right)dx \right]\]

\[\text{Putting }\tan x = t\]

\[ \Rightarrow \sec^2 x dx = dt\]

\[\text{When }x \to 0; t \to 0\]

\[\text{and }x \to \frac{\pi}{2}; t \to \infty \]

\[ \therefore 2I = 2\pi \int_0^\frac{\pi}{2} \frac{dt}{a^2 + b^2 t^2}\]

\[ \Rightarrow I = \frac{\pi}{b^2} \int_0^\frac{\pi}{2} \frac{dt}{\frac{a^2}{b^2} + t^2}\]

\[ = \frac{\pi}{b^2} \times \frac{b}{a} \left[ \tan^{- 1} \left( \frac{bt}{a} \right) \right]_0^\infty \]

\[ = \frac{\pi}{ab}\left[ \frac{\pi}{2} - 0 \right]\]

\[ = \frac{\pi}{ab} \times \frac{\pi}{2}\]

\[ = \frac{\pi^2}{2ab} \]

\[\text{Hence }I = \frac{\pi^2}{2ab}\]