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

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

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

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

CanadianAsian
 one year 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.
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