Question

Show that ${\int }_0^{\pi /2}\frac{x}{\sin x+\cos x}dx=\frac{\pi }{2\sqrt{2}}\ln (\sqrt{2}+1)$

Answer 1

Let

$I={\int }_0^{\frac{\pi }{2}}\frac{x}{\sin x+\cos x}dx$

$I={\int }_0^{\frac{\pi }{2}}\frac{x}{\sin x+\cos x}dx={\int }_0^{\frac{\pi }{2}}\frac{\frac{\pi }{2}-x}{\sin x+\cos x}dx$

$\Rightarrow I=\frac{\pi }{2}{\int }_0^{\frac{\pi }{2}}\frac{dx}{\sin x+\cos x}-I$

$\Rightarrow I=\frac{\pi }{4}{\int }_0^{\frac{\pi }{2}}\frac{x}{\sin x+\cos x}dx=\frac{\pi }{4}{\int }_0^{\frac{\pi }{2}}\frac{dx}{\frac{2\tan \frac{x}{2}}{1+\tan x^2\frac{x}{2}}+\frac{1-\tan x^2\frac{x}{2}}{1+\tan x^2\frac{x}{2}}}$

$\Rightarrow I=\frac{\pi }{4}{\int }_0^{\frac{\pi }{2}}\frac{\left(1+\tan x^2\frac{x}{2}\right)}{1+2\tan \frac{x}{2}-{\tan }^2\frac{x}{2}}dx$

$=\frac{\pi }{4}{\int }_0^{\frac{\pi }{2}}\frac{{\sec }^{2}x}{1+2\tan \frac{x}{2}-{\tan }^2\frac{x}{2}}dx$

$Let{\tan }^2\frac{x}{2}=t$

$\Rightarrow {\sec }^2\frac{x}{2}dx=2dt$

So,

$limitsx\to 0\Rightarrow t\to 0$

$x\to \frac{\pi }{2}\Rightarrow t\to 1$

$I=\frac{\pi }{4}{\int }_0^1\frac{2dt}{1+2t-t^2}=\frac{-\pi }{2}{\int }_0^1\frac{2dt}{\left(t-1-\sqrt{2}\right)\left(t-1+\sqrt{2}\right)}$

$=\frac{-\pi }{2}{\int }_0^1\frac{1}{2\sqrt{2}}\left[\frac{1}{t-1-\sqrt{2}}-\frac{1}{t-1-\sqrt{2}}\right]dt$

$=\frac{-\pi }{4\sqrt{2}}{\int }_0^1\frac{dt}{\left(t-1-\sqrt{2}\right)}+\frac{-\pi }{4\sqrt{2}}{\int }_0^1\frac{dt}{\left(t-1+\sqrt{2}\right)}$

$=\frac{-\pi }{4\sqrt{2}}{\left[\ln \left|\left(t-1-\sqrt{2}\right)\right|\right]}_0^1+\frac{\pi }{4\sqrt{2}}{\left[\ln \left|\left(t-1+\sqrt{2}\right)\right|\right]}_0^1$

$=\frac{\pi }{4\sqrt{2}}\left[\ln \sqrt{2}-\ln \left(1+\sqrt{2}\right)\right]+\left[\ln \sqrt{2}-\ln \left(\sqrt{2}-1\right)\right]$

$=\frac{\pi }{4\sqrt{2}}\left[\ln \sqrt{2}+\ln \sqrt{2}+\ln \left(1+\sqrt{2}\right)-\ln \left(\sqrt{2}-1\right)\right]$

$=\frac{\pi }{4\sqrt{2}}\ln \frac{\left(1+\sqrt{2}\right)}{\left(\sqrt{2}-1\right)}$

$=\frac{\pi }{4\sqrt{2}}\ln {\left(1+\sqrt{2}\right)}^2$

$=\frac{\pi }{2\sqrt{2}}\ln \left(\sqrt{2}+1\right)$

Hence, the following question shown.