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siddarth95
 one year ago
Best ResponseYou've already chosen the best response.0\[\int\limits_{0}^{\pi/2}\frac{ dx }{ \sin x + \cos x } = \frac{ 1 }{ \sqrt{2} } \ln \ \frac{ \sqrt{2}+1 }{ \sqrt{2}1 }\]

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

Jonask
 one year ago
Best 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

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

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

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

Meepi
 one year 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

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

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

siddarth95
 one year ago
Best ResponseYou've already chosen the best response.0its all good .. thanks for the responses :)

experimentX
 one year 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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