Question

Integration of 1/(1+sinx) limit π/4 to 3π/4

Answer 1

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Answer 2

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$\int \frac{1}{1 + sinx } dx= \int ({sec}^{2} x - secx \: tanx \: )dx$

Breaking the integral :
$= \int {sec}^{2} x \: dx \: - \int secx \: tanx \: dx$

Hence,
$= tan x \: - secx$

Now, limiting from :
$\frac{\pi}{4} - \frac{3\pi}{4}$
∆ We get the desired value as :

$( \tan( \frac{3\pi}{4} ) - \sec( \frac{3\pi }{4} ) ) - ( \tan( \frac{\pi}{4} ) - \sec( \frac{\pi }{4} ) ) \\ \\ = ( - 1 + { \sqrt{2} } ) - (1 - \sqrt{2} ) = 2( \sqrt{2}-1 )$

Hence, the desired integral limit is : 2( √2 - 1 )