Question

${\int }_0^{\frac{\pi }{2}}\frac{sinxcosx}{1+si{n}^{4}x}dx=$

• A. $\frac{\pi }{2}$
• B. $\frac{\pi }{4}$
• C. $\frac{\pi }{6}$
• D. $\frac{\pi }{8}$

Answer 1

The correct option is D $\frac{\pi }{8}$

$Putsi{n}^{2}x=t\Rightarrow dt=2sinxcosxdx$
$Now{\int }_0^{\frac{\pi }{2}}\frac{sinxcosx}{1+si{n}^{4}x}dx=\frac{1}{2}{\int }_0^1\frac{1}{1+t^2}dt=\frac{1}{2}[ta{n}^{-1}t{]}_0^1=\frac{\pi }{8}$