Question

If $I={\int }_0^{\pi /2}{(\sqrt{\sin x}+\sqrt{\cos x})}^{-4}dx$, then $I$ equals

• A. $1$
• B. $1/2$
• C. $1/3$
• D. $1/4$

Answer 1

The correct option is C $1/3$

$I={\int }_0^{\pi /2}{(\sqrt{\sin x}+\sqrt{\cos x})}^{-4}dx={\int }_0^{\pi /2}\frac{1}{{\cos }^{2}x{(\sqrt{\tan x}+1)}^4}dx$
Substitute $\tan x=t^2$
$I={\int }_0^1\frac{2t}{{(t+1)}^4}dt=2{\int }_0^1(\frac{1}{{(t+1)}^3}-\frac{1}{{(t+1)}^4})dt$
$={[-4\frac{1}{{(t+1)}^2}+6\frac{1}{{(t+1)}^3}]}_0^1=\frac{1}{3}$