Question

$\int \frac{\tan 2\theta }{\sqrt{{\cos }^6\theta +{\sin }^6\theta }}d\theta$

Answer 1

${\cos }^6\theta +{\sin }^6\theta =\frac{1+3{\cos }^2\theta }{4}$

$\int \frac{\tan 2\theta }{\sqrt{{\cos }^6\theta +{\sin }^6\theta }}d\theta =\int \frac{2\tan 2\theta }{\sqrt{1+3{\cos }^{2}2\theta }}d\theta =\int \frac{2\sin 2\theta }{\cos 2\theta \sqrt{1+3{\cos }^{2}2\theta }}d\theta$

put $t=\cos 2\theta ⟹dt=-2\sin 2\theta d\theta$

$⟹\int \frac{-1}{t\sqrt{1+3t^2}}dt=\int \frac{1}{\sqrt{\frac{1}{3t^2}+1}}\times \frac{-dt}{\sqrt{3}{t}^2}$

put $l=\frac{1}{\sqrt{3}t}⟹dl=\frac{-dt}{\sqrt{3}{t}^2}$

$⟹\int \frac{dl}{\sqrt{l^2+1}}=\ln (l+\sqrt{l^2+1})+C=\ln (\frac{\text{sec}2\theta +\sqrt{{\text{sec}}^{2}2\theta +3}}{3})+C$