Question

\[\int_0^\frac{\pi}{2} \frac{\cos x}{\left( \cos\frac{x}{2} + \sin\frac{x}{2} \right)^n}dx\]

Answer 1

\[\text{Let I }=\int_0^\frac{\pi}{2} \frac{\cos x}{\left( \cos\frac{x}{2} + \sin\frac{x}{2} \right)^n}dx\]

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

Put

\[\cos\frac{x}{2} + \sin\frac{x}{2} = z\]

\[\therefore \left( - \sin\frac{x}{2} \times \frac{1}{2} + \cos\frac{x}{2} \times \frac{1}{2} \right)dx = dz\]
\[ \Rightarrow \left( \cos\frac{x}{2} - \sin\frac{x}{2} \right)dx = 2dz\]

When

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

When

\[x \to \frac{\pi}{2}, z \to \sqrt{2} ..................\left( z = \cos\frac{\pi}{4} + \sin\frac{\pi}{4} = \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}} = \frac{2}{\sqrt{2}} = \sqrt{2} \right)\]

\[\therefore I = 2 \int_1^\sqrt{2} \frac{dz}{z^{n - 1}}\]
\[ = \left.2 \times \frac{z^{2 - n}}{2 - n}\right|_1^\sqrt{2} \]
\[ = \frac{2}{\left( 2 - n \right)}\left[ \left( \sqrt{2} \right)^{2 - n} - 1 \right]\]
\[ = \frac{2}{\left( 2 - n \right)}\left( 2^\frac{2 - n}{2} - 1 \right)\]
\[ = \frac{2}{\left( 2 - n \right)}\left( 2^{1 - \frac{n}{2}} - 1 \right)\]