Evaluate $\int\limits_{\pi/3}^{\pi/2} \frac{\sqrt{1 + \cos x}}{(1 - \cos x)^{5/2}} \, dx$

$\frac{3}{2}$

The correct answer is Option (1) → $\frac{3}{2}$

Let $I = \int\limits_{\pi/3}^{\pi/2} \frac{\sqrt{1 + \cos x}}{(1 - \cos x)^{5/2}} \, dx = \int\limits_{\pi/3}^{\pi/2} \frac{\left( 2\cos^2 \frac{x}{2} \right)^{1/2}}{\left( 2\sin^2 \frac{x}{2} \right)^{5/2}} \, dx$

$∵\cos x = 1 - 2\sin^2 \frac{x}{2}$

$\cos x = 2\cos^2 \frac{x}{2} - 1$

$= \frac{\sqrt{2}}{4\sqrt{2}} \int\limits_{\pi/3}^{\pi/2} \frac{\cos\left(\frac{x}{2}\right)}{\sin^5\left(\frac{x}{2}\right)} \, dx = \frac{1}{4} \int\limits_{\pi/3}^{\pi/2} \frac{\cos\left(\frac{x}{2}\right)}{\sin^5\left(\frac{x}{2}\right)} \, dx$

Put $\sin \frac{x}{2} = t$

$\Rightarrow \cos \frac{x}{2} \cdot \frac{1}{2} \, dx = dt$

$\Rightarrow \cos \frac{x}{2} \, dx = 2 \, dt$

As $x \to \frac{\pi}{3}$, then $t \to \frac{1}{2}$

and $x \to \frac{\pi}{2}$, then $t \to \frac{1}{\sqrt{2}}$

$∴I = \frac{2}{4} \int\limits_{1/2}^{1/\sqrt{2}} \frac{dt}{t^5} = \frac{1}{2} \left[ \frac{t^{-5+1}}{-5+1} \right]_{1/2}^{1/\sqrt{2}}$

$= \frac{1}{2} \left[ \frac{t^{-4}}{-4} \right]_{1/2}^{1/\sqrt{2}} \Rightarrow \frac{-1}{8} \left[ \frac{1}{t^4} \right]_{1/2}^{1/\sqrt{2}}$

$= -\frac{1}{8} \left[ \frac{1}{\left(\frac{1}{\sqrt{2}}\right)^4} - \frac{1}{\left(\frac{1}{2}\right)^4} \right]$

$= -\frac{1}{8} (4 - 16) = \frac{12}{8} = \frac{3}{2}$