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  1. #1
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    Question integration

    The integral (from 0 to pi/2) of (costheta)(sintheta) d(theta)

    Thank you very much.
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by kittycat View Post
    The integral (from 0 to pi/2) of (costheta)(sintheta) d(theta)

    Thank you very much.
    remember your identities

    \int_{0}^{\frac {\pi}{2}} \cos \theta \sin \theta ~d \theta = \frac {1}{2} \int_{0}^{\frac {\pi}{2}} \sin 2 \theta~d \theta which is a simple integration by substitution problem

    or you can use substitution directly. let u = \sin \theta
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  3. #3
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    Question what about this integration?

    hi Jhevon,

    What about this integration....?

    the integral (from 0 to pi/2) of (cos^2 theta) (sintheta) d(theta)

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  4. #4
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by kittycat View Post
    hi Jhevon,

    What about this integration....?

    the integral (from 0 to pi/2) of (cos^2 theta) (sintheta) d(theta)

    Thank you very much.
    substitution: u = \cos \theta


    (remember, if you're having trouble with the integrals even after i tell you the substitutions, don't be afraid to say so)
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  5. #5
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    Question

    hi Jhevon,

    Thank you very much for your kindness.

    Actually the question is (8/4^a) times the integral (from 0 to pi/2) of (cos^2 theta) (sintheta) d(theta) times the integral (from 0 to a) r^4 dr.

    The answer is (8a)/15.

    I have trouble to get to this answer. If you could help, please show me how you can arrive this answer. Thank you very much.
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  6. #6
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by kittycat View Post
    hi Jhevon,

    Thank you very much for your kindness.

    Actually the question is (8/4^a) times the integral (from 0 to pi/2) of (cos^2 theta) (sintheta) d(theta) times the integral (from 0 to a) r^4 dr.

    The answer is (8a)/15.

    I have trouble to get to this answer. If you could help, please show me how you can arrive this answer. Thank you very much.
    are you sure 8/(4^a) shouldn't be 8/(a^4) ??
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  7. #7
    is up to his old tricks again! Jhevon's Avatar
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    you've gone silent on me. assuming it should have been a^4 instead of 4^a, we proceed as follows:

    \frac {8}{a^4} \int_{0}^{ \frac {\pi}{2}} \cos^2 \theta \sin \theta~d \theta \int_{0}^{a} r^4~dr = \frac {8}{a^4} \left[ - \frac {1}{3} \cos^3 \theta \right]_{0}^{ \frac {\pi}{2}} \left[ \frac {1}{5}r^5 \right]_{0}^{a}

    (as i said earlier, i used the substitution u = \cos \theta for the first integral)

    ........................................... = \frac {8}{a^4} \left[ \frac {1}{3} \right] \left[ \frac {a^5}{5} \right]

    ........................................... = \frac {8a}{15}
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  8. #8
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    You can also use the general result.
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  9. #9
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    Hi Jhevon,

    Thank you very much.

    I am sorry because I encountered trouble in re-connecting my internet.

    Yes, you are right. It should be 8/a^4

    Thank you very much.
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