Question

Evaluate
$I={\int }_0^{\pi /2}\frac{1}{5\csc +3\sin x}dx$

Answer 1

We have,

$I={\int }_0^{\frac{\pi }{2}}\frac{dx}{5\cos ecx+3\sin x}I={\int }_0^{\frac{\pi }{2}}\frac{\sin x}{5+3{\sin }^{2}x}dx={\int }_0^{\frac{\pi }{2}}\frac{\sin xdx}{5+3\left(1-{\cos }^{2}x\right)}I={\int }_0^{\frac{\pi }{2}}\frac{\sin xdx}{8-3{\cos }^{2}x}$

$Lett=\cos x$

$\frac{dt}{dx}=-\sin x$

Therefore,

$I={\int }_1^0\frac{-dt}{8-3t^2}={\int }_0^1\frac{dt}{8-3t^2}I=\frac{1}{3}{\int }_0^1\frac{dt}{\frac{8}{3}-t^2}=\frac{1}{3}{\int }_0^1\frac{dt}{{\left(\sqrt{\frac{8}{3}}\right)}^2-t^2}=\frac{1}{3}\frac{1}{2\sqrt{\frac{8}{3}}}{\left[\ln \left(\frac{\sqrt{\frac{8}{3}}+t}{\sqrt{\frac{8}{3}}-t}\right)\right]}_0^1=\frac{\sqrt{3}}{3\times 2\sqrt{8}}\left[\ln \left(\frac{\sqrt{\frac{8}{3}}+1}{\sqrt{\frac{8}{3}}-1}\right)-\ln 1\right]=\frac{1}{2\sqrt{3}\sqrt{8}}\ln \left(\frac{\sqrt{8}+\sqrt{3}}{\sqrt{8}-\sqrt{3}}\right)=\frac{1}{2\sqrt{24}}\ln \left(\frac{\sqrt{8}+\sqrt{3}}{\sqrt{8}-\sqrt{3}}\right)$

Hence, this is the answer.