Got Homework?
Connect with other students for help. It's a free community.
Here's the question you clicked on:
 0 viewing
What is the integral of
2sinx/1+cos^2x? I can't seem to get it :(
 one year ago
 one year ago
What is the integral of 2sinx/1+cos^2x? I can't seem to get it :(
 one year ago
 one year ago

This Question is Closed

jotopia34Best ResponseYou've already chosen the best response.0
\[\int\limits_{0}^{\pi/6}2sinx/1+\cos^2x\]
 one year ago

jotopia34Best ResponseYou've already chosen the best response.0
I used u=cosx, but i got a weird quotient
 one year ago

hartnnBest ResponseYou've already chosen the best response.0
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 ?
 one year ago

jotopia34Best ResponseYou've already chosen the best response.0
yes harnn i did but now can i just pull out the 2?
 one year ago

jotopia34Best ResponseYou've already chosen the best response.0
oh pellet, i think i see it, the bottom part is inverse tan right?
 one year ago

CanadianAsianBest ResponseYou've already chosen the best response.2
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.
 one year ago
See more questions >>>
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.