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siddarth95Best ResponseYou've already chosen the best response.0
\[\int\limits_{0}^{\pi/2}\frac{ dx }{ \sin x + \cos x } = \frac{ 1 }{ \sqrt{2} } \ln \ \frac{ \sqrt{2}+1 }{ \sqrt{2}1 }\]
 one year ago

JonaskBest ResponseYou've already chosen the best response.1
let u=sin x +cos x du=cosxsinx dx
 one year ago

JonaskBest ResponseYou've already chosen the best response.1
\[\int \frac{1}{ \cos x+\sin x}dx\] multiply by cos xsin x both numerator and denominator
 one year ago

JonaskBest ResponseYou've already chosen the best response.1
\[\int \frac{\cos x\sin x}{\cos^2x\sin^2x} dx=\int \frac{\cancel{\cos x\sin x}du}{\cos 2x}\frac{du}{\cancel{\cos x\sin x}}\]
 one year ago

JonaskBest ResponseYou've already chosen the best response.1
\[\int \frac{du}{\cos 2x}=\int \sec 2x=\ln\sec 2x+\tan 2x\]/2
 one year ago

JonaskBest ResponseYou've already chosen the best response.1
error i never substituted u
 one year ago

MeepiBest ResponseYou've already chosen the best response.0
I 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
 one year ago

MeepiBest ResponseYou've already chosen the best response.0
Pretty sure others can help you if you get stuck :3
 one year ago

experimentXBest ResponseYou've already chosen the best response.0
dw:1362839531405:dw
 one year ago

siddarth95Best ResponseYou've already chosen the best response.0
its all good .. thanks for the responses :)
 one year ago

experimentXBest ResponseYou've already chosen the best response.0
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 likedw:1362841238898:dw
 one year ago
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