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What is the integral of 2sinx/1+cos^2x? I can't seem to get it :(

Calculus1
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\[\int\limits_{0}^{\pi/6}2sinx/1+\cos^2x\]
I used u=cosx, but i got a weird quotient
when u=cos x du =-sin x dx so, your new integral will be \(\int \dfrac{-2}{1+u^2}du\) did you got this ?

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Other answers:

yes harnn i did but now can i just pull out the -2?
oh pellet, i think i see it, the bottom part is inverse tan right?
This is u substitution. Say that \[u = \cos x\] Then the derivative of u is: \[du = -sinxdx\] Then change your bounds: \[\cos(0) = 1\] which is your lower bound and your upper bound is: \[\cos(\Pi/6) = \frac{ \sqrt3 }{ 2 }\] So then you can substitute these into the equation to get \[\int\limits_{1}^{\frac{\sqrt3}{2}}\frac{ -2 }{ 1+u^2 }\] which is the same as \[-2\int\limits_{1}^{\frac{\sqrt3}{2}}\frac{ 1 }{ 1+u^2 }\] and \[\int\limits_{}^{}\frac{ 1 }{ 1 + u^2 }\] is just arctan. So just take \[-2*(\arctan(\frac{ \sqrt3 }{ 2 }) - \arctan(1))\] To get your answer.
yes.
thank u very much!!

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