Question

Solve:$\underset{\frac{\pi }{4}}{\overset{\frac{3\pi }{4}}{\int }}\frac{\pi }{1+\sin x}dx$

Answer 1

$={\int }_{\frac{\pi }{4}}^{\frac{3\pi }{4}}\frac{\pi }{1+\sin x}dx={\int }_{\frac{\pi }{4}}^{\frac{3\pi }{4}}\frac{\pi (1-\sin x)}{(1+\sin x)(1-\sin x)}dx$

$={\int }_{\frac{\pi }{4}}^{\frac{3\pi }{4}}\frac{(1-\sin x)}{{\cos }^{2}x}dx$

$={\int }_{\frac{\pi }{4}}^{\frac{3\pi }{4}}({\sec }^{2}x-\tan x\sec x)dx$

$=\pi [\tan x-\sec x{]}_{\frac{\pi }{4}}^{\frac{3\pi }{4}}$

$=\pi [-1+\sqrt{2}-1+\sqrt{2}]$

$=\pi [2\sqrt{2}-2]$

$=2\pi (\sqrt{2}-1)$