Question

$\underset{\mathrm{\pi }/3}{\overset{\mathrm{\pi }/2}{\int }}\frac{\sqrt{1+\cos x}}{{\left(1-\cos x\right)}^{3/2}}dx$

Answer 1

$$\begin{gathered} {\int }_{\frac{\mathrm{\pi }}{3}}^{\frac{\mathrm{\pi }}{2}}\frac{\sqrt{1+\cos x}}{{\left(1-\cos x\right)}^{{\displaystyle \frac{3}{2}}}}dx \\ ={\int }_{\frac{\mathrm{\pi }}{3}}^{\frac{\mathrm{\pi }}{2}}\frac{\sqrt{1+\cos x}}{{\left(1-\cos x\right)}^{{\displaystyle \frac{3}{2}}}}\times \frac{\sqrt{1-\cos x}}{\sqrt{1-\cos x}}dx \\ ={\int }_{\frac{\mathrm{\pi }}{3}}^{\frac{\mathrm{\pi }}{2}}\frac{\sqrt{1-{\cos }^{2}x}}{{\left(1-\cos x\right)}^2}dx \\ ={\int }_{\frac{\mathrm{\pi }}{3}}^{\frac{\mathrm{\pi }}{2}}\frac{\sin x}{{\left(1-\cos x\right)}^2}dx \\ \mathrm{Let}1-\cos x=t,\mathrm{Then}\sin xdx=dt \\ \mathrm{When}x=\frac{\mathrm{\pi }}{3},t=\frac{1}{2}\mathrm{and}x=\frac{\mathrm{\pi }}{2},t=1 \\ \mathrm{Therefore}\mathrm{the}\mathrm{integral}\mathrm{becomes} \\ ={\int }_{\frac{1}{2}}^1\frac{\mathrm{dt}}{{\mathrm{t}}^2} \\ ={\left[-\frac{1}{\mathrm{t}}\right]}_{\frac{1}{2}}^1 \\ =-1+2 \\ =1 \end{gathered}$$