Question

${\int }_0^{\pi /4}\frac{\sin x+\cos x}{{\cos }^{2}x+{\sin }^{4}x}$

Answer 1

$I={\int }_0^{x/4}\frac{\sin x+\cos x}{{\cos }^{2}x+{\sin }^{4}x}dx$

$={\int }_0^{x/4}\frac{4(\sin x+\cos x)}{2.2{\cos }^{2}x+{\left(2{\sin }^{2}x\right)}^2}dx$

$=4{\int }_0^{x/4}\frac{\sin x+\cos x}{2(1+\cos x2){(1-\cos x2)}^2}dx$

$=4{\int }_0^{x/4}\frac{\sin x+\cos x}{3+{\cos }^{2}2x}=4{\int }_0^{x/4}\frac{\sin x+\cos x}{4-{\sin }^{2}2x}dx$

Let $\sin x-\cos x=z\Rightarrow (\cos x+\sin x)dx=dz$

and $z^2={(\sin x-\cos x)}^2\Rightarrow z^2=1-\sin 2x$

Or, $\sin 2x=1-z^2$

$I=4{\int }_0^{x/4}\frac{\sin x+\cos x}{4-{\sin }^{2}2x}dx$

$=4{\int }_{-1}^0\frac{dz}{2^2-{(1-z^2)}^2}={\int }_{-1}^0\frac{4dz}{(3-z^2)(z^2+1)}$

$={\int }_{-1}^0[\frac{1}{z^2+1}+\frac{1}{{3-z}^2}]dz$

$={[{\tan }^{-1}z+\frac{1}{2\sqrt{3}}\log \frac{\sqrt{3}+z}{\sqrt{3}-z}]}_{-1}^0$

$=0-[{\tan }^{-1}(-1)+\frac{1}{2\sqrt{3}}\log \frac{\sqrt{3}-1}{\sqrt{3}+1}]$

$=\frac{\pi }{4}+\frac{1}{2\sqrt{3}}\log \frac{\sqrt{3}+1}{\sqrt{3}-1}$

$=\frac{\pi }{4}+\frac{1}{\sqrt{3}}\log \frac{\sqrt{3}+1}{\sqrt{2}}$Ans