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

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anonymous
  • anonymous
\[\int \frac{\cos x-\sin x}{\cos^2x-\sin^2x} dx=\int \frac{\cancel{\cos x-\sin x}du}{\cos 2x}\frac{du}{\cancel{\cos x-\sin x}}\]
anonymous
  • anonymous
\[\int \frac{du}{\cos 2x}=\int \sec 2x=\ln|\sec 2x+\tan 2x|\]/2
anonymous
  • anonymous
error i never substituted u
anonymous
  • anonymous
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
anonymous
  • anonymous
Pretty sure others can help you if you get stuck :3
experimentX
  • experimentX
|dw:1362839531405:dw|
anonymous
  • anonymous
its all good .. thanks for the responses :)
experimentX
  • experimentX
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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