Mathematics - Definite Integration Question with Solution | TestHub

${\int }_0^{\pi /3}\frac{\tan x}{\sqrt{2k\sec x}}dx$

$=\frac{1}{\sqrt{2k}} {\int }_0^{\pi /3}\frac{\sin x}{\sqrt{\cos x}}dx$

Put$t=\cos x$

$dt= -\sin x dx,$also$x=0, t=1;x=\frac{\pi }{3}, t=\frac{1}{2}$

$\therefore \frac{1}{\sqrt{2k}} \underset{1}{\overset{1/2}{\int }}\frac{-dt}{\sqrt{t}}$

$=\frac{1}{\sqrt{2k}}{\underset{1/2}{\overset{1}{\int }}{t}^{-\frac{1}{2} }dt=\frac{1}{\sqrt{2k}}\times 2 {\left[t^{\frac{1}{2}}\right]}_{1/2}^1}_{ }$

$=\frac{\sqrt{2}}{\sqrt{k} }\left(1-\frac{1}{\sqrt{2}}\right)=\frac{\sqrt{2}-1}{\sqrt{k}}$

Given$\frac{\sqrt{2}-1}{\sqrt{k}}=1-\frac{1}{\sqrt{2}}=\frac{\sqrt{2}-1}{\sqrt{2}}$

$\therefore k=2$