Question

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

Answer 1

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

Answer 2

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

Answer 3

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 ?

Answer 4

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

Answer 5

oh pellet, i think i see it, the bottom part is inverse tan right?

Answer 6

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.

Answer 7

yes.

Answer 8

thank u very much!!