Question

Integration help again....
integrate (sin2x)/(1+cos^2x)
http://www.wolframalpha.com/input/?i=integrate+%28sin2x%29%2F%281%2Bcos%5E2x%29

please explain how was du manipulated so that if fits into the equation in step 2 @_@

Answer 1

I got this :D

\[sin(2x)=2sinxcosx\]

This is your double angle formular

\[du = -2sinxcosx\]

Answer 2

Move the - over so it's a -du and then replace 2sinxcosx with sin(2x)

Answer 3

damn i forgot about those xD i only thought of the basic trig identities...thanks a lot dude!

Answer 4

\[-du=2sinxcosx\]
\[-du=sin(2x)\]

I was just doing some of these myself recently for a summer Calculus class I'm in, so yeah I recognized it immediately. :D

Answer 5

i'm taking up integral calculus now, and i did not expect to encounter some trig substitution problems >_< i hated proving identities in trigonometry lol

Answer 6

\[\cos^2x=\frac{1}{2}(1+\cos2x)\]\[\int\limits{\frac{\sin(2x)}{\frac{1}{2}+\frac{1}{2}\cos(2x)+1}}dx=\int\limits{\frac{\sin(2x)}{\frac{3}{2}+\frac{1}{2}\cos(2x)}}dx\][=\int\limits{\frac{2\sin(2x)}{3+\cos(2x)}}dx\];
let u=3+cos(2x)
du=-2sin(2x)dx
so the integral becomes\[\int\limits{\frac{-du}{u}}=-\ln(u)+C\]so\[\int\limits{\frac{\sin(2x)}{1+\cos(2x)}}dx=-\ln(3+\cos(2x))\]

Answer 7

Thats another way to think about it :)

Answer 8

the third line should be\[=\int\limits\limits{\frac{2\sin(2x)}{3+\cos(2x)}}dx\]

Answer 9

@lalaly is that the half angle identity? ugh everything becomes confusing when you apply trigonometry into cacluclus

Answer 10

yes it is ... its not confusing , u just want to make it look easier to apply substitution

Answer 11

the part where you think of what to substitute is xD integration itself is easier hahaha

Answer 12

lol integration is fun :P