Question

\[\int_0^\frac{\pi}{2} \frac{\cos^2 x}{1 + 3 \sin^2 x}dx\]

Answer 1

\[Let\ I = \int_0^\frac{\pi}{2} \frac{\cos^2 x}{1 + 3 \sin^2 x}dx\]
\[ = \int_0^\frac{\pi}{2} \frac{\cos^2 x}{1 + 3\left( 1 - \cos^2 x \right)}dx\]
\[ = \int_0^\frac{\pi}{2} \frac{\cos^2 x}{4 - 3 \cos^2 x}dx\]
\[ = - \frac{1}{3} \int_0^\frac{\pi}{2} \frac{4 - 3 \cos^2 x - 4}{4 - 3 \cos^2 x}dx\]

\[= - \frac{1}{3} \int_0^\frac{\pi}{2} dx + \frac{4}{3} \int_0^\frac{\pi}{2} \frac{1}{4 - 3 \cos^2 x}dx\]
\[ = \left.- \frac{1}{3} x\right|_0^\frac{\pi}{2} + \frac{4}{3} \int_0^\frac{\pi}{2} \frac{\sec^2 x}{4 \sec^2 x - 3}dx ..............\left( \text{Dividing numerator and denominator by} \cos^2 x \right)\]
\[ = - \frac{1}{3}\left( \frac{\pi}{2} - 0 \right) + \frac{4}{3} \int_0^\frac{\pi}{2} \frac{\sec^2 x}{4\left( 1 + \tan^2 x \right) - 3}dx\]
\[ = - \frac{\pi}{6} + \frac{4}{3} \int_0^\frac{\pi}{2} \frac{\sec^2 x}{4 \tan^2 x + 1}dx\]

Put tanx = z

\[\therefore \sec^2 xdx = dz\]

When

\[x \to 0, z \to 0\]

When

\[x \to \frac{\pi}{2}, z \to \infty\]

\[\therefore I = - \frac{\pi}{6} + \frac{4}{3} \int_0^\infty \frac{dz}{4 z^2 + 1}\]
\[ = - \frac{\pi}{6} + \frac{4}{3} \int_0^\infty \frac{dz}{\left( 2z \right)^2 + 1}\]
\[ = \left.- \frac{\pi}{6} + \frac{4}{3} \times \frac{\tan^{- 1} 2z}{2}\right|_0^\infty \]
\[ = - \frac{\pi}{6} + \frac{2}{3}\left( \tan^{- 1} \infty - \tan^{- 1} 0 \right)\]
\[ = - \frac{\pi}{6} + \frac{2}{3}\left( \frac{\pi}{2} - 0 \right)\]
\[ = - \frac{\pi}{6} + \frac{\pi}{3}\]
\[ = \frac{\pi}{6}\]