Question

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.

Answer 1

You can try using integration by substitution. Let \[y = \sin^2x\]

Answer 2

How am I gonna do it with the two intervals given?

Answer 3

Do it without the intervals first. Then insert the intervals

Answer 4

put y=x/2
so dy=dx/2
that is dx=2dy\[\int\limits_{}^{}2\csc^2(y) dy=-2 \cot(y)=-2\cot(x/2)=F(x)\]
\[\int\limits_{\pi/2}^{2\pi/3}2\csc^2(y) dy=F(upperlimit)- F(lowerlimit)\]

Answer 5

Thank you again for your help. :) I really appreciate this.

Answer 6

@Annand: Thank you for showing me how to do the solution. :)

Answer 7

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. :(

Answer 8

\[\tanπ/3 = \sqrt{3}\]
\[\tanπ/4 = 1\]
2-2√3

Answer 9

I mean 1/2 - 1/2*√3/3
\[\frac{1}{2} \left[ 1 - \frac{\sqrt{3}}{3} \right]\]

Answer 10

@ wengzie:
the answer is
\[-2[cot(\pi/3)-cot(\pi/4)]\]
where
\[\cot(\pi/3)=1/\sqrt(3) , \cot(\pi/4)=1\]