A community for students.
Here's the question you clicked on:
 0 viewing

This Question is Closed

jotopia34
 2 years ago
Best ResponseYou've already chosen the best response.0\[\int\limits_{0}^{\pi/6}2sinx/1+\cos^2x\]

jotopia34
 2 years ago
Best ResponseYou've already chosen the best response.0I used u=cosx, but i got a weird quotient

hartnn
 2 years ago
Best ResponseYou've already chosen the best response.0when u=cos x du =sin x dx so, your new integral will be \(\int \dfrac{2}{1+u^2}du\) did you got this ?

jotopia34
 2 years ago
Best ResponseYou've already chosen the best response.0yes harnn i did but now can i just pull out the 2?

jotopia34
 2 years ago
Best ResponseYou've already chosen the best response.0oh pellet, i think i see it, the bottom part is inverse tan right?

CanadianAsian
 2 years ago
Best ResponseYou've already chosen the best response.2This 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.
Ask your own question
Sign UpFind more explanations on OpenStudy
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.