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## How to integrate $\int_{\pi /3}^{\pi /2} {\frac{{\sqrt {1 + \cos x} }}{{{{(1 – \cos x)}^{5/2}}}}} dx$ ~~~~~[NCERT Exemp. Q. 41,Page 166 ]

# Let $I = \int_{\pi /3}^{\pi /2} {\frac{{\sqrt {1 + \cos x} }}{{{{(1 – \cos x)}^{5/2}}}}} dx$

$ = \int_{\pi /3}^{\pi /2} {\frac{{\sqrt {1 + \cos x} }}{{{{(1 – \cos x)}^2}\sqrt {1 + \cos x} }}} dx$

$ = \int_{\pi 3}^{\pi 2} {\frac{1}{{\left( {1 – {{\cos }^2}x} \right)}}} dx = \int_{\pi 3}^{\pi 2} {\frac{1}{{{{\sin }^2}x}}} dx$

$ = \int_{\pi 3}^{\pi 2} {{{{\mathop{\rm cosec}\nolimits} }^2}} xdx = [ – \cot x]_{\pi /3}^{\pi /2}$

$ = – \left[ {\cot \frac{\pi }{2} – \cot \frac{\pi }{3}} \right] = – \left[ {0 – \frac{1}{{\sqrt 3 }}} \right] = + \frac{1}{{\sqrt 3 }}$

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$ = \int_{\pi /3}^{\pi /2} {\frac{{\sqrt {1 + \cos x} }}{{{{(1 – \cos x)}^2}\sqrt {1 + \cos x} }}} dx$

$ = \int_{\pi 3}^{\pi 2} {\frac{1}{{\left( {1 – {{\cos }^2}x} \right)}}} dx = \int_{\pi 3}^{\pi 2} {\frac{1}{{{{\sin }^2}x}}} dx$

$ = \int_{\pi 3}^{\pi 2} {{{{\mathop{\rm cosec}\nolimits} }^2}} xdx = [ – \cot x]_{\pi /3}^{\pi /2}$

$ = – \left[ {\cot \frac{\pi }{2} – \cot \frac{\pi }{3}} \right] = – \left[ {0 – \frac{1}{{\sqrt 3 }}} \right] = + \frac{1}{{\sqrt 3 }}$