Question

${\int }_0^{\pi /2}\frac{xdx}{\sin x+\cos x}=$

• A. $\frac{\pi }{2\sqrt{2}}\log (\sqrt{2}+1)$
• B. $\frac{{\pi }^2}{2\sqrt{2}}\log (\sqrt{2}+1)$
• C. $\pi \log (\sqrt{2}+1)$
• D. $\log (\sqrt{2}+1)$

Answer 1

The correct option is A $\frac{\pi }{2\sqrt{2}}\log (\sqrt{2}+1)$

Let $I={\int }_0^{\frac{\pi }{2}}\frac{xdx}{sinx+cosx}$ ...(1)

Using property ${\int }_a^{b}f\left(x\right)dx={\int }_a^{b}f(a+b-x)dx$

We get

$I={\int }_0^{\frac{\pi }{2}}\frac{(\frac{\pi }{2}-x)dx}{sin(\frac{\pi }{2}-x)+cos(\frac{\pi }{2}-x)}={\int }_0^{\frac{\pi }{2}}\frac{(\frac{\pi }{2}-x)dx}{cosx+sinx}$ ...(2)

Adding (1) and (2), we get

$2I=\frac{\pi }{2}{\int }_0^{\frac{\pi }{2}}\frac{dx}{sinx+cosx}$

Substituting $tan\left(\frac{x}{2}\right)=t\Rightarrow \frac{1}{2}{sec}^2\left(\frac{x}{2}\right)dx=dt$

We get $sinx=\frac{2t}{1+t^2},cosx=\frac{1-t^2}{1+t^2},dx=\frac{2dt}{1+t^2}$

Therefore

$2I=\frac{\pi }{2}{\int }_0^1\frac{2dt}{(1+t^2)(\frac{2t}{1+t^2}+\frac{1-t^2}{1+t^2})}=\pi {\int }_0^1\frac{dt}{-t^2+2t+1}=\pi {\int }_0^1\frac{dt}{2-{(t-1)}^2}$

Substituting $t-1=s\Rightarrow dt=ds$

$2I=\pi {\int }_{-1}^0\frac{ds}{2-s^2}=\pi {\int }_{-1}^0\frac{ds}{2(1-\frac{s^2}{2})}$

Again substituting $\frac{s}{\sqrt{2}}=u\Rightarrow ds=\sqrt{2}du$

$2I=\frac{\pi }{\sqrt{2}}{\int }_{-1/\sqrt{2}}^0\frac{du}{1-u^2}=\frac{\pi }{\sqrt{2}}{\left[\frac{1}{2}\log \left(\frac{u+1}{1-u}\right)\right]}_{-1\sqrt{2}}^0=\frac{\pi }{2\sqrt{2}}\log (\sqrt{2}+1)$