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wengzie

  • 3 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.

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  1. Kira_Yamato
    • 3 years ago
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    You can try using integration by substitution. Let \[y = \sin^2x\]

  2. wengzie
    • 3 years ago
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    How am I gonna do it with the two intervals given?

  3. Kira_Yamato
    • 3 years ago
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    Do it without the intervals first. Then insert the intervals

  4. Annand
    • 3 years ago
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    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
    • 3 years ago
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    Thank you again for your help. :) I really appreciate this.

  6. wengzie
    • 3 years ago
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    @Annand: Thank you for showing me how to do the solution. :)

  7. wengzie
    • 3 years ago
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    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
    • 3 years ago
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    \[\tanπ/3 = \sqrt{3}\] \[\tanπ/4 = 1\] 2-2√3

  9. Kira_Yamato
    • 3 years ago
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    I mean 1/2 - 1/2*√3/3 \[\frac{1}{2} \left[ 1 - \frac{\sqrt{3}}{3} \right]\]

  10. Annand
    • 3 years ago
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    @ wengzie: the answer is \[-2[cot(\pi/3)-cot(\pi/4)]\] where \[\cot(\pi/3)=1/\sqrt(3) , \cot(\pi/4)=1\]

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