Question

${\int }_{\pi /3}^{\pi /2}\frac{\sqrt{1+cosx}}{(1-cosx{)}^{5/2}}dx$

Answer 1

${\int }_{\pi /3}^{\pi /2}\frac{\sqrt{1+cosx}}{(1-cosx{)}^{5/2}}dxLetI={\int }_{\pi /3}^{\pi /2}\frac{\sqrt{1+cosx}}{(1-cosx{)}^{5/2}}dx={\int }_{\pi /3}^{\pi /2}\frac{\sqrt{1+cosx}}{(1-cosx{)}^2\sqrt{1+cosx}}dx={\int }_{\pi /3}^{\pi /2}\frac{1}{(1-co{s}^{2}x)}dx=={\int }_{\pi /3}^{\pi /2}\frac{1}{si{n}^{2}x}dx={\int }_{\pi /3}^{\pi /2}cose{c}^{2}xdx=[-cotx{]}_{\pi /3}^{\pi /2}=-[cot\frac{\pi }{2}-cot\frac{\pi }{3}]=-[0-\frac{1}{\sqrt{3}}]=+\frac{1}{\sqrt{3}}AlternateMethodLetI={\int }_{\pi /3}^{\pi /2}\frac{\sqrt{1+cosx}}{(1-cosx{)}^5/2}dx={\int }_{\pi /3}^{\pi /2}\frac{{\left(2co{s}^2\frac{x}{2}\right)}^{1/2}}{{\left(2si{n}^2\frac{x}{2}\right)}^5/2}dx=\frac{\sqrt{2}}{4\sqrt{2}}={\int }_{\pi /3}^{\pi /2}\frac{cos\left(\frac{x}{2}\right)}{si{n}^5\left(\frac{x}{2}\right)}dx=\frac{1}{4}{\int }_{\pi /3}^{\pi /2}\frac{cos\left(\frac{x}{2}\right)}{si{n}^5\left(\frac{x}{2}\right)}dxPutsin\frac{x}{2}=t\Rightarrow cos\frac{x}{2}.\frac{1}{2}dx=dt\Rightarrow cos\frac{x}{2}dx=2dtAsx\to \frac{\pi }{3},thent\to \frac{1}{2}andx\to \frac{\pi }{2},thent\to \frac{1}{\sqrt{2}}\therefore I=\frac{2}{4}{\int }_{1/2}^{1\sqrt{2}}\frac{dt}{t^5}=\frac{1}{2}{\left[\frac{t^{-5+1}}{-5+1}\right]}_{1/2}^{1/\sqrt{2}}=-\frac{1}{8}[\frac{1}{{\left(\frac{1}{\sqrt{2}}\right)}^4}-\frac{1}{{\left(\frac{1}{2}\right)}^4}]=-\frac{1}{8}(4-16)=\frac{12}{8}=\frac{3}{2}$

Note If we integrate the trigonometric function in different ways [using different identities] then, we can get different answers.