# Evaluate the following:$\int\limits_{\Large \frac{\pi}{3}}^{\Large \frac{\pi}{2}}\large\frac{\sqrt{1+\cos x}}{(1-\cos x)^{\large\frac{5}{2}}}$

Toolbox:
• $1+\cos x=2 \cos ^2\large\frac{x}{2}$
• $1-\cos x=2 \sin^2 \large\frac{x}{2}$
• $\int udv=uv-\int vdu$
• $\log a +\log b=\log(ab)$
Step 1:
Let $I=\int \limits_{\Large\frac{\pi}{2}}^{\Large\frac{\pi}{2}} \large \frac{\sqrt {1+\cos x}}{(1-\cos x)^{\Large\frac{5}{2}}}$
Since $1+\cos x=2 \cos ^2 \large\frac{x}{2}$
$\sqrt {1+\cos x}=\sqrt {2} \cos \large\frac{x}{2}$
Similarly $(1-\cos x)=2 \sin ^2 \large\frac{x}{2}$
$(1-\cos x)^{\Large\frac{5}{2}}=(2)^{\Large\frac{5}{2}}.\sin ^5 \large\frac{x}{2}$
$\qquad\qquad\;\;\;\;=4\sqrt 2.\sin ^5 \large\frac{x}{2}$
Hence $I=\int \limits_{\Large\frac{\pi}{3}}^{\Large\frac{\pi}{2}} \large\frac{\sqrt 2 \cos\Large\frac {x}{2}}{4 \sqrt 2 \sin ^5 \Large\frac{x}{2}}=\large\frac{1}{4} \int \limits_{\Large\frac{\pi}{3}}^{\Large\frac{\pi}{2}} \frac{\cos\Large\frac {x}{2}}{\sin ^5\Large\frac {x}{2}}$
Let $\sin \large\frac{x}{2}$$=t on differentiating with respect to x we get, \large\frac{1}{2}$$ \cos \large\frac{x}{2} $$dx=dt =>\cos \large\frac{x}{2}$$dx=2 dt$
When we substitute for x, the limits also change
As $x\to \large\frac{\pi}{3}$$;t\to \large\frac{1}{2}$$\quad[\sin \large\frac{\pi}{6}=\large\frac{1}{2}]$
As $x\to \large\frac{\pi}{2}$$;t\to \large\frac{1}{\sqrt 2}$$\quad[\sin \large\frac{\pi}{4}=\large\frac{1}{\sqrt 2}]$
Hence on substituting we get,
$I=\large\frac{1}{4}$$.2 \int \limits_{\Large\frac{1}{2}}^{\Large\frac{1}{\sqrt 2}} \large\frac{dt}{t^5} Step 2: On integrating we get, I= \large\frac{1}{2} \bigg[\frac{\Large t^{-5+1}}{-5+1}\bigg]_{\Large\frac{1}{2}}^{\Large\frac{1}{\sqrt 2}} =\large\frac{-1}{8} \bigg[\frac{1}{t^4}\bigg]_{\Large\frac{1}{2}}^{\Large\frac{1}{\sqrt 2}} On applying the limits we get, I=\large\frac{-1}{8}\bigg[\bigg(\frac{1}{\frac{1}{\sqrt 2}}\bigg)^4-\bigg(\frac{1} {\frac{1}{2}}\bigg)^4\bigg] =\large\frac{-1}{8}$$[4-16]$
$=\large\frac{12}{8}=\frac{3}{2}$