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Pls help me. How do you integrate this problem, it has two intervals: pi/2 below and 2pi/3 above the integral sign. The equation is csc^2 (x/2) dx. Thank you so much for any help.
 2 years ago
 2 years ago
Pls help me. How do you integrate this problem, it has two intervals: pi/2 below and 2pi/3 above the integral sign. The equation is csc^2 (x/2) dx. Thank you so much for any help.
 2 years ago
 2 years ago

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Kira_YamatoBest ResponseYou've already chosen the best response.0
You can try using integration by substitution. Let \[y = \sin^2x\]
 2 years ago

wengzieBest ResponseYou've already chosen the best response.0
How am I gonna do it with the two intervals given?
 2 years ago

Kira_YamatoBest ResponseYou've already chosen the best response.0
Do it without the intervals first. Then insert the intervals
 2 years ago

AnnandBest ResponseYou've already chosen the best response.1
put y=x/2 so dy=dx/2 that is dx=2dy\[\int\limits_{}^{}2\csc^2(y) dy=2 \cot(y)=2\cot(x/2)=F(x)\] \[\int\limits_{\pi/2}^{2\pi/3}2\csc^2(y) dy=F(upperlimit) F(lowerlimit)\]
 2 years ago

wengzieBest ResponseYou've already chosen the best response.0
Thank you again for your help. :) I really appreciate this.
 2 years ago

wengzieBest ResponseYou've already chosen the best response.0
@Annand: Thank you for showing me how to do the solution. :)
 2 years ago

wengzieBest ResponseYou've already chosen the best response.0
can i ask what the final answer is? so i could check my answer. i'm not very sure if i got the answer. i thought the answer should not be an equation, and i got an eqution. :(
 2 years ago

Kira_YamatoBest ResponseYou've already chosen the best response.0
\[\tanπ/3 = \sqrt{3}\] \[\tanπ/4 = 1\] 22√3
 2 years ago

Kira_YamatoBest ResponseYou've already chosen the best response.0
I mean 1/2  1/2*√3/3 \[\frac{1}{2} \left[ 1  \frac{\sqrt{3}}{3} \right]\]
 2 years ago

AnnandBest ResponseYou've already chosen the best response.1
@ wengzie: the answer is \[2[cot(\pi/3)cot(\pi/4)]\] where \[\cot(\pi/3)=1/\sqrt(3) , \cot(\pi/4)=1\]
 2 years ago
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