Question

Solve: ${\int }_0^{\frac{\pi }{2}}\frac{\sin x}{1+{\cos }^{2}x}dx$

Answer 1

${\int }_0^{\pi /2}\frac{sinxdx}{1+co{s}^{2}x}$

Let cos x =t ; x=0, t=1

-sin x dx= dt $x=\pi /2,t=0$

${\int }_1^0\frac{-dt}{1+t^2}$

$={\int }_0^1\frac{dt}{1+t^2}$ $\left[{\int }_a^{b}f\right(x)dx=-{\int }_b^{a}f(x\left)dx\right]$

$=ta{n}^{-1}\left(t\right){|}_0^1$

$=\frac{\pi }{4}+0$

$=\frac{\pi }{4}$

$\therefore {\int }_0^{\pi /2}\frac{sinxdx}{1+co{s}^{2}x}=\frac{\pi }{4}$