## wengzie 4 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.

1. Kira_Yamato

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

2. wengzie

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

3. Kira_Yamato

Do it without the intervals first. Then insert the intervals

4. Annand

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

5. wengzie

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

6. wengzie

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

7. wengzie

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. :(

8. Kira_Yamato

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

9. Kira_Yamato

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

10. Annand

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