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\[\int\limits_{0}^{\pi/2}\frac{ dx }{ \sin x + \cos x } = \frac{ 1 }{ \sqrt{2} } \ln \ \frac{ \sqrt{2}+1 }{ \sqrt{2}-1 }\]
let u=sin x +cos x du=cosx-sinx dx
\[\int \frac{1}{ \cos x+\sin x}dx\] multiply by cos x-sin x both numerator and denominator

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Other answers:

\[\int \frac{\cos x-\sin x}{\cos^2x-\sin^2x} dx=\int \frac{\cancel{\cos x-\sin x}du}{\cos 2x}\frac{du}{\cancel{\cos x-\sin x}}\]
\[\int \frac{du}{\cos 2x}=\int \sec 2x=\ln|\sec 2x+\tan 2x|\]/2
error i never substituted u
I don't have time to put an explanation here, but use the substitution u = tan(x/2):
Pretty sure others can help you if you get stuck :3
its all good .. thanks for the responses :)
there's one another way to do it .. change all trigs into half angles, and change sines and cosines into tan and sec .. you should end up something like|dw:1362841238898:dw|

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