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Integral substitution? Not getting the right answer?
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\[\int\limits_{}^{} \frac{ \sin 2x }{ 1+\cos^2x } dx\]
When I am doing it, I'm letting \[u=1+\cos^2x\]
and what are you getting stuck on?
I'm getting \[\frac{ du }{ dx }= -2sinxcosx\] I don't think that will help me find the integral
it will if you recall your trig identities .... what is another way to write sin(2x)?
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2sinxcosx! Oh that makes more sense
:) you should have the rest of it then
Yup I do thanks!
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