## jotopia34 2 years ago What is the integral of 2sinx/1+cos^2x? I can't seem to get it :(

1. jotopia34

$\int\limits_{0}^{\pi/6}2sinx/1+\cos^2x$

2. jotopia34

I used u=cosx, but i got a weird quotient

3. hartnn

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 ?

4. jotopia34

yes harnn i did but now can i just pull out the -2?

5. jotopia34

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.

7. hartnn

yes.

8. jotopia34

thank u very much!!