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anonymous
 3 years ago
Help prove this integral
anonymous
 3 years ago
Help prove this integral

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anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\[\int\limits_{0}^{\pi/2}\frac{ dx }{ \sin x + \cos x } = \frac{ 1 }{ \sqrt{2} } \ln \ \frac{ \sqrt{2}+1 }{ \sqrt{2}1 }\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0let u=sin x +cos x du=cosxsinx dx

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\[\int \frac{1}{ \cos x+\sin x}dx\] multiply by cos xsin x both numerator and denominator

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\[\int \frac{\cos x\sin x}{\cos^2x\sin^2x} dx=\int \frac{\cancel{\cos x\sin x}du}{\cos 2x}\frac{du}{\cancel{\cos x\sin x}}\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\[\int \frac{du}{\cos 2x}=\int \sec 2x=\ln\sec 2x+\tan 2x\]/2

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0error i never substituted u

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I don't have time to put an explanation here, but use the substitution u = tan(x/2): http://wwwmath.mit.edu/~djk/18_01/chapter24/section03.html

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Pretty sure others can help you if you get stuck :3

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.0dw:1362839531405:dw

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0its all good .. thanks for the responses :)

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.0there'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 likedw:1362841238898:dw
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