Question

${\int }_0^{\pi /2}\frac{\cos xdx}{1-{\sin }^{2}x+{\sin }^{4}x}=\frac{\pi }{k}+\frac{1}{4\sqrt{\left(3\right)}}log\left(\frac{(2+\sqrt{3})}{(2-\sqrt{3})}\right)$. Find the value of $k$.

Answer 1

${\int }_0^{\pi /2}\frac{\cos xdx}{1-{\sin }^{2}x+{\sin }^{4}x}$
Put $\sin x=t$
$\cos xdx=dt$
When $x=0\Rightarrow t=0$
When $x=\frac{\pi }{2}\Rightarrow t=1$
$I={\int }_0^1\frac{dt}{t^4-t^2+1}$
$I=\frac{1}{2}{\int }_0^1\frac{\frac{2}{t^2}dt}{t^2+\frac{1}{t^2}-1}$
$I=\frac{1}{2}{\int }_0^1\frac{\frac{1}{t^2}+1+\frac{1}{t^2}-1dt}{t^2+\frac{1}{t^2}+2-2-1}$
$I=\frac{1}{2}{\int }_0^1\frac{\frac{1}{t^2}+1dt}{(t-\frac{1}{t}{)}^2+(1{)}^2}-\frac{1}{2}{\int }_0^1\frac{1-\frac{1}{t^2}dt}{(t+\frac{1}{t}{)}^2-(\sqrt{3}{)}^2}$
Put $t-\frac{1}{t}=u$ in first integral
$\Rightarrow (1+\frac{1}{t^2})dt=du$
Put $t+\frac{1}{t}=v$ in second integral
$\Rightarrow (1-\frac{1}{t^2})dt=dv$
$I=\frac{1}{2}{\int }_{-\infty }^0\frac{du}{u^2+1^2}+\frac{1}{2}{\int }_2^{\infty }\frac{dv}{v^2-(\sqrt{3}{)}^2}$
$I=\frac{1}{2}[{\tan }^{-1}u{]}_{-\infty }^0+\frac{1}{4\sqrt{3}}[\log |\frac{v-\sqrt{3}}{v+\sqrt{3}}|{]}_2^{\infty }$
$I=\frac{\pi }{4}+\frac{1}{4\sqrt{3}}\log |\frac{2+\sqrt{3}}{2-\sqrt{3}}|$