Question

Help prove this integral

Answer 1

\[\int\limits_{0}^{\pi/2}\frac{ dx }{ \sin x + \cos x } = \frac{ 1 }{ \sqrt{2} } \ln \ \frac{ \sqrt{2}+1 }{ \sqrt{2}-1 }\]

Answer 2

let u=sin x +cos x
du=cosx-sinx dx

Answer 3

\[\int \frac{1}{ \cos x+\sin x}dx\]
multiply by cos x-sin x both numerator and denominator

Answer 4

\[\int \frac{\cos x-\sin x}{\cos^2x-\sin^2x} dx=\int \frac{\cancel{\cos x-\sin x}du}{\cos 2x}\frac{du}{\cancel{\cos x-\sin x}}\]

Answer 5

\[\int \frac{du}{\cos 2x}=\int \sec 2x=\ln|\sec 2x+\tan 2x|\]/2

Answer 6

error i never substituted u

Answer 7

I don't have time to put an explanation here, but use the substitution u = tan(x/2):
http://www-math.mit.edu/~djk/18_01/chapter24/section03.html

Answer 8

Pretty sure others can help you if you get stuck :3