Question

Solve : ${\int }_0^{\pi /2}\frac{xdx}{sinx+cosx}$

Answer 1

${\int }_0^{\Pi /2}\frac{x}{sinx+cosx}dx$

Say,

$I={\int }_0^{\Pi /2}\frac{x}{sinx+cosx}dx⟶\left(1\right)$

$\Rightarrow I={\int }_0^{\Pi /2}\frac{\Pi /2-x}{\sin (\frac{\Pi }{2}-x)+\cos (\frac{\Pi }{2}-x)}dx$ [Using ${\int }_0^{a}f\left(x\right)dx={\int }_0^{a}f(a-x)dx$]

$\Rightarrow I={\int }_0^{\Pi /2}\frac{\Pi /2-x}{cosx+sinx}dx⟶\left(2\right)$

Adding $\left(1\right)$ and $\left(2\right)$ we get

$\Rightarrow 2I={\int }_0^{\Pi /2}\frac{\Pi /2}{sinx+cosx}dx$

$\Rightarrow I=\frac{\Pi }{4}{\int }_0^{\Pi /2}\frac{1}{sinx+cosx}dx$

$\Rightarrow I=\frac{\Pi }{4}{\int }_0^{\Pi /2}\frac{1}{\sqrt{2}(sinx.\frac{1}{\sqrt{2}}+cosx.\frac{1}{\sqrt{2}})}dx$

$\Rightarrow I=\frac{\Pi }{4\sqrt{2}}{\int }_0^{\Pi /2}\frac{1}{(cosx.\cos \frac{\Pi }{4}+sinx.\sin \frac{\Pi }{4})}dx$

$\Rightarrow I=\frac{\Pi }{4\sqrt{2}}{\int }_0^{\Pi /2}\frac{1}{\cos (x-\frac{\Pi }{4})}dx$

$\Rightarrow I=\frac{\Pi }{4\sqrt{2}}{\int }_0^{\Pi /2}\sec (x-\frac{\Pi }{4})dx$

$\Rightarrow I=\frac{\Pi }{4\sqrt{2}}log|\sec (x-\frac{\Pi }{4})+\tan (x-\frac{\Pi }{4})|{]}_0^{\Pi /2}$

$\Rightarrow I=\frac{\Pi }{4\sqrt{2}}[log|\sec \frac{\Pi }{4}+\tan \frac{\Pi }{4}|-log|\sec \frac{\Pi }{4}+\tan \frac{-\Pi }{4}|]$

$\Rightarrow I=\frac{\Pi }{4\sqrt{2}}[log(\sqrt{2}+1)-log(\sqrt{2}-1)]$ $[\sec \Pi /4=\sqrt{2}=\sec (-\Pi /4)]$

$\Rightarrow I=\frac{\Pi }{4\sqrt{2}}log\left(\frac{\sqrt{2}+1}{\sqrt{2}-1}\right)$ $[\tan \Pi /4=1;\tan (-\Pi /4)=-1]$