Question

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

Answer 1

$I={\int }_0^{\pi /2}\frac{{\sin }^{2}x}{\sin x+\cos x}dx$
$I={\int }_0^{\pi /2}\frac{\frac{1-\cos 2x}{2}}{\sin x+\cos x}dx[\because \cos 2x=1-2{\sin }^{2}x]$
$I=\frac{1}{2}{\int }_0^{\pi /2}\frac{dx}{\sin x+\cos x}-\frac{1}{2}{\int }_0^{\pi /2}\frac{\cos 2x}{\sin x+\cos x}dx$
$P=\frac{1}{2}{\int }_0^{\pi /2}\frac{dx}{\sin x+\cos x},Q=\frac{1}{2}{\int }_0^{\pi /2}\frac{\cos 2x}{\sin x+\cos x}dx$
$\therefore I=P-Q⟶1$
$Q=\frac{1}{2}{\int }_0^{\pi /2}\frac{\cos 2x}{\sin x+\cos x}dx=\frac{1}{2}{\int }_0^{\pi /2}\frac{{\cos }^{2}x-{\sin }^{2}x}{\sin x+\cos x}dx$
$Q=\frac{1}{2}{\int }_0^{\pi /2}\frac{(\cos x-\sin x)(\cos x+\sin x)}{\sin x+\cos x}dx$
$Q=\frac{1}{2}{\int }_0^{\pi /2}(\cos x-\sin x)dx$
$Q=\frac{1}{2}{|\sin x+\cos x|}_0^{\pi /2}$
$Q=\frac{1}{2}[1+0-0-1]=0⟶2$
$P=\frac{1}{2}{\int }_0^{\pi /2}\frac{dx}{\sin x+\cos x}$
$P=\frac{1}{2}{\int }_0^{\pi /2}\frac{dx}{2\sin \frac{x}{2}\cos \frac{x}{2}+1-2{\sin }^2\frac{x}{2}}[\because \sin 2\theta =2\sin \theta \cos \theta ,\cos 2\theta =1-2{\sin }^2\theta ]$
divide numerator and denominator by ${\cos }^2\frac{x}{2},$ we get
$P=\frac{1}{2}{\int }_0^{\pi /2}\frac{{\sec }^2\frac{x}{2}dx}{2\tan \frac{x}{2}+{\sec }^2\frac{x}{2}-2{\tan }^2\frac{x}{2}}$
$P=\frac{1}{2}{\int }_0^{\pi /2}\frac{{\sec }^2\frac{x}{2}dx}{2\tan \frac{x}{2}+1-{\tan }^2\frac{x}{2}}[\because {\sec }^2\frac{x}{2}-{\tan }^2\frac{x}{2}=1]$
Put $\tan \frac{x}{2}=t⟹\frac{1}{2}{\sec }^2\frac{x}{2}dx=dt$
when $x=0⟹t=0$ & $x=\frac{\pi }{2}⟹t=1$
$P={\int }_0^1\frac{dt}{2t+1-t^2}={\int }_0^1\frac{dt}{-(t^2+1-2t)+2}$
$P={\int }_0^1\frac{dt}{{\left(\sqrt{2}\right)}^2-{(t-1)}^2}$
$P=\frac{1}{2\sqrt{2}}{\left[\log \left|\frac{\sqrt{2}+t-1}{\sqrt{2}-t+1}\right|\right]}_0^1$
$[\because \int \frac{1}{a^2-x^2}dx=\frac{1}{2a}\log \left|\frac{a+x}{a-x}\right|]$
$P=\frac{1}{2\sqrt{2}}[\log \left|\frac{\sqrt{2}+t-1}{\sqrt{2}-t+1}\right|-\log \left|\frac{\sqrt{2}+0-1}{\sqrt{2}-0+1}\right|]$
$P=\frac{1}{2\sqrt{2}}[\log 1-\log \left|\frac{{(\sqrt{2}-1)}^2}{2-1}\right|]$
$[\because \frac{\sqrt{2}-1}{\sqrt{2}+1}=\frac{\sqrt{2}-1}{\sqrt{2}+1}\times \frac{\sqrt{2}-1}{\sqrt{2}-1}]$
$P=\frac{-1}{2\sqrt{2}}\log [2+1-2\sqrt{2}]$$P=\frac{-1}{2\sqrt{2}}\log [3-2\sqrt{2}]$$P=\frac{1}{2\sqrt{2}}\log [\frac{1}{3-2\sqrt{2}}\times \frac{3+2\sqrt{2}}{3+2\sqrt{2}}]$
$P=\frac{1}{2\sqrt{2}}\log \left[\frac{3+2\sqrt{2}}{9-8}\right]$
$P=\frac{1}{2\sqrt{2}}\log (3+2\sqrt{2})⟶3$
$I=P-Q=\frac{1}{2\sqrt{2}}\log (3+2\sqrt{2})$