Question

\[\int_0^\frac{\pi}{2} \sqrt{\cos x - \cos^3 x}\left( \sec^2 x - 1 \right) \cos^2 xdx\]

Answer 1

\[\text{Let I }=\int_0^\frac{\pi}{2} \sqrt{\cos x - \cos^3 x}\left( \sec^2 x - 1 \right) \cos^2 xdx\]

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

Put `cos x = z^2`

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

When

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

When

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

\[\therefore I = - \int_1^0 z\left( 1 - z^4 \right)2zdz\]
\[ = - 2 \int_1^0 z^2 dz + 2 \int_1^0 z^6 dz\]
\[ = \left.- 2 \times \frac{z^3}{3}\right|_1^0 + \left.2 \times \frac{z^7}{7}\right|_1^0 \]
\[ = - \frac{2}{3}\left( 0 - 1 \right) + \frac{2}{7}\left( 0 - 1 \right)\]
\[ = \frac{2}{3} - \frac{2}{7}\]
\[ = \frac{8}{21}\]