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integrate sinx/(1+sinx)^1/2 dx
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http://www.wolframalpha.com/input/?i=integrate++sinx%2F%281%2Bsinx%29%5E1%2F2+dx click show steps.
use\[\sin^2x+\cos^2x=1\] after rearranging to become \[\cos^2x=1-\sin^2x\] \[\cos x=\sqrt{1-\sin^2x}\]
so you are integrating sin x /cos x which is the same as integrating tanx
A better way would be to substitute \(x = \frac{\pi}2 - 2y\). Then you can use multiple angle formula to get rid of that square-root.
idk. if it's better though
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But that's how I do it, generally.
but i can not remember such formulas i can only remember sin^2x +cos^2x=1
co2x = cos^2x - \sin^2x => 1 + cos2x = 2cos^2x
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