Question

\[\int_{- \frac{\pi}{2}}^\frac{\pi}{2} \frac{- \frac{\pi}{2}}{\sqrt{\cos x \sin^2 x}}dx\]

Answer 1

\[Let I = \int_{- \frac{\pi}{2}}^\frac{\pi}{2} \frac{- \frac{\pi}{2}}{\sqrt{\cos x \sin^2 x}}dx\]
\[ = - \frac{\pi}{2} \int_{- \frac{\pi}{2}}^\frac{\pi}{2} \frac{1}{\sqrt{\cos x \sin^2 x}}dx\]
\[ = - \frac{\pi}{2} \int_{- \frac{\pi}{2}}^\frac{\pi}{2} \frac{1}{\sqrt{\cos x}\left| \sin x \right|}dx\]
\[ = - \frac{\pi}{2} \times 2 \int_0^\frac{\pi}{2} \frac{1}{\sqrt{\cos x}\left| \sin x \right|}dx .................\left[ f\left( - x \right) = \sqrt{\cos\left( - x \right)}\left| \sin\left( - x \right) \right| = \sqrt{\cos x}\left| - \sin x \right| = \sqrt{\cos x}\left| \sin x \right| = f\left( x \right) \right]\]

\[= - \pi \int_0^\frac{\pi}{2} \frac{1}{\sqrt{\cos x}\sin x}dx ...................\left( \left| \sin x \right| = \sin x, 0 \leq x \leq \frac{\pi}{2} \right)\]
\[ = - \pi \int_0^\frac{\pi}{2} \frac{\sin x}{\sqrt{\cos x}\left( 1 - \cos^2 x \right)}dx\]

Put `cosx = z^2`

\[\therefore - \sin x\ dx = 2zdz\]

When

\[x \to 0, z \to 1\]

When

\[x \to \frac{\pi}{2}, z \to 0\]

\[\therefore I = 2\pi \int_1^0 \frac{zdz}{z\left( 1 - z^4 \right)}\]
\[ = 2\pi \int_1^0 \frac{dz}{1 - z^4}\]
\[ = 2\pi \int_1^0 \frac{dz}{\left( 1 - z \right)\left( 1 + z \right)\left( 1 + z^2 \right)}\]

Now,

\[\frac{1}{\left( 1 - z \right)\left( 1 + z \right)\left( 1 + z^2 \right)} = \frac{A}{1 - z} + \frac{B}{1 + z} + \frac{Cz + D}{1 + z^2}\]
\[ \Rightarrow 1 = A\left( 1 + z \right)\left( 1 + z^2 \right) + B\left( 1 - z \right)\left( 1 + z^2 \right) + \left( Cz + D \right)\left( 1 - z \right)\left( 1 + z \right)\]

Putting z = 1, we get

\[A = \frac{1}{4}\]

Putting z = −1, we get

\[B = \frac{1}{4}\]

Putting z = 0, we get
\[1 = A + B + D\]
\[ \Rightarrow D = 1 - \frac{1}{4} - \frac{1}{4} = \frac{1}{2}\]

Equating coefficient of z³ on both sides, we get

\[A - B + C = 0\]
\[ \Rightarrow \frac{1}{4} - \frac{1}{4} + C = 0\]
\[ \Rightarrow C = 0\]

\[\therefore I = 2\pi \int_1^0 \frac{dz}{\left( 1 - z \right)\left( 1 + z \right)\left( 1 + z^2 \right)}\]
\[ = 2\pi \int_1^0 \frac{\frac{1}{4}}{1 - z}dz + 2\pi \int_1^0 \frac{\frac{1}{4}}{1 + z}dz + 2\pi \int_1^0 \frac{\frac{1}{2}}{1 + z^2}dz\]
\[ = \left.\frac{2\pi}{4} \times \frac{\log\left( 1 - z \right)}{- 1}\right|_1^0 + \left.\frac{2\pi}{4} \times \log\left( 1 + z \right)\right|_1^0 + \left.\frac{2\pi}{2} \times \tan^{- 1} z\right|_1^0 \]
\[ = - \frac{\pi}{2}\left( \log1 - \log0 \right) + \frac{\pi}{2}\left( \log1 - \log2 \right) + \pi\left( \tan^{- 1} 0 - \tan^{- 1} 1 \right)\]
\[ = - \frac{\pi}{2}\left[ 0 - \left( - \infty \right) \right] + \frac{\pi}{2}\left( 0 - \log2 \right) + \pi\left( 0 - \frac{\pi}{4} \right)\]
\[ = - \infty - \frac{\pi}{2}\log2 - \frac{\pi^2}{4}\]
\[ = - \infty\]