Question

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

Answer 1

$I={\int }_0^{\pi /2}\frac{si{n}^{2}x}{1+sinx+cosx}dx...\left(1\right)$

$I={\int }_0^{\pi /2}\frac{si{n}^2(\frac{\pi }{2}-x)}{1+sin(\frac{\pi }{2}-x)+cos(\frac{\pi }{2}-x)}dx={\int }_0^{\pi /2}\frac{co{s}^{2}x}{1+cosx+sinx}...\left(2\right)$

$\left[{\int }_0^{a}f\right(x)dx={\int }_0^{a}f(a-x\left)dx\right]$

Adding (1), (2)

$\Rightarrow 2I={\int }_0^{\pi /2}\frac{dx}{1+cosx+sinx}={\int }_0^{\pi /2}\frac{dx}{1+\frac{1-ta{n}^2\frac{x}{2}}{1+ta{n}^2\frac{x}{2}}+\frac{2tan\frac{x}{2}}{1+ta{n}^2\frac{x}{2}}}$

$={\int }_0^{\pi /2}\frac{se{c}^2\frac{x}{2}dx}{2(tan\frac{x}{2}+1)}$

Let $tan\frac{x}{2}+1=u\Rightarrow du=se{c}^2\frac{x}{2}dx$

when $x=0\Rightarrow u=1,x=\frac{\pi }{2},u=2$

$\Rightarrow 2I={\int }_0^2\frac{du}{u}=[\log u{]}_0^2=\log 2$

$I=\frac{1}{2}\log 2$