Question

Integrate ${\int }_{-\frac{\pi }{4}}^{\frac{\pi }{4}}\frac{1}{1-\sin x}dx$.

Answer 1

$I={\int }_{-\pi /4}^{\pi /4}\frac{1}{1-\sin x}dx$

$I={\int }_{-\pi /4}^{\pi /4}\frac{1}{1-\sin (\frac{\pi }{4}-(-\frac{\pi }{4})-x)}dx={\int }_{-\pi /4}^{\pi /4}\frac{1}{1+\sin x}dx$

$2I={\int }_{-\pi /4}^{\pi /4}(\frac{1}{1-\sin x}+\frac{1}{1+\sin x})dx={\int }_{-\pi /4}^{\pi /4}\frac{2}{\cos x}dx$

$\Rightarrow I={\int }_{-\pi /4}^{\pi /4}\sec xdx=\log |\sec x+\tan x|{]}_{-\pi /4}^{\pi /4}$

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

$=\log |3+2\sqrt{2}|$