Question

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

Answer 1

$\begin{array}{c}{\int }_0^{\pi /2}\frac{1}{\sin x+\cos x}dx\\ {\int }_0^{\pi /2}\frac{{\sec }^{2}x}{2\tan \frac{\pi }{2}+1-{\tan }^2\frac{\pi }{2}}dx\end{array}$

${\int }_0^{\pi /2}\frac{{\sec }^2\frac{x}{2}}{1-\left({\tan }^2\frac{\pi }{2}-2{\tan }^2\frac{\pi }{2}\right)}dx$

Put $\tan \frac{x}{2}=tatx=0,\tan \frac{\pi }{2}=0=t_{lowerlimit}$

$\frac{1}{2}{\sec }^2\frac{x}{2}=dtx=\frac{\pi }{2},\tan \frac{x}{2}=1=t_{upperlimit}$

$\left\{use\left|\frac{dx}{a^2-x^2}=\frac{k}{2a}n!\left|\frac{a+x}{a-x}\right|\right|\right\}$

$=\frac{2}{2\sqrt{2}}n!{\left|\frac{\sqrt{2}+\left(t-1\right)}{\sqrt{2}-\left(t-1\right)}\right|}_0^1$

$=\frac{1}{\sqrt{2}}n!\left|\frac{\sqrt{2}-1}{\sqrt{2}+1}\right|$