Question

If $I=636{\int }_0^{\frac{\pi }{4}}{(\sqrt{\sin x}+\sqrt{\cos x})}^{-4}dx$ then $I$ equals

Answer 1

$I=636{\int }_0^{\frac{\pi }{4}}{(\sqrt{\sin x}+\sqrt{\cos x})}^{-4}dx=636{\int }_0^{\frac{\pi }{4}}\frac{1}{{\cos }^{2}x{(\sqrt{\tan x}+1)}^4}dx$
Substituting

$\tan x=t^2\Rightarrow {\sec }^{2}xdx=2tdt$
$I=636{\int }_0^1\frac{2t}{{(1+t)}^4}dt=1272{\int }_0^1(\frac{1}{{(1+t)}^3}-\frac{1}{{(1+t)}^4})dt=1272{[-\frac{1}{2{(1+t)}^2}+\frac{1}{{3(1+t)}^3}]}_0^1=106$