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Math Help - [SOLVED] Simple trig integration

  1. #1
    Junior Member Evales's Avatar
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    [SOLVED] Simple trig integration

    Can you guys help. I haven't done this stuff in a while.

    How would you integrate:
    (sin x)^2

    Thanks if you can shed light.
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  2. #2
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    Quote Originally Posted by Evales View Post
    Can you guys help. I haven't done this stuff in a while.

    How would you integrate:
    (sin x)^2

    Thanks if you can shed light.
    Recall the double angle formula \cos (2x) = 1 - 2 \sin^2 x .....
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  3. #3
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Evales View Post
    Can you guys help. I haven't done this stuff in a while.

    How would you integrate:
    (sin x)^2

    Thanks if you can shed light.
    note that \sin^2 x = \frac {1 - \cos 2x}2. integrate that. it is easier

    EDIT: Thanks a lot, Mr Fantastic!
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  4. #4
    Flow Master
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    Quote Originally Posted by Jhevon View Post
    note that \sin^2 x = \frac {1 - \cos 2x}2. intgrate that. it is easier
    What a dynamic duo ..... Note how posts #2 and #3 just seamlessly flow
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  5. #5
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by mr fantastic View Post
    What a dynamic duo ..... Note how posts #2 and #3 just seamlessly flow
    Yup, can't help it, being around the Flow Master and all
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  6. #6
    Junior Member Evales's Avatar
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    Never heard of that particular identity before.

    I assume that in order to do this I shouldn't use identities that aren't on the syllabus. Anyway. Getting past that. How would I do it? Should I use substitution or the chain rule ect.
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  7. #7
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by Evales View Post
    Never heard of that particular identity before.

    I assume that in order to do this I shouldn't use identities that aren't on the syllabus. Anyway. Getting past that. How would I do it? Should I use substitution or the chain rule ect.
    \cos(2x)=\cos^2(x)-\sin^2(x)
    from that we can replace

    \cos^2(x)=1-\sin^2(x)

    to get

    \cos(2x)=1-2\sin^2(x)

    From there you can find what they said

    so

    \int\sin^2(x)dx=\int\frac{1-\cos(2x)}{2}dx=\frac{x}{2}-\frac{1}{2}\int\cos(2x)dx

    for the last one isolate the derivative of the quanity and be done with it =D
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  8. #8
    Junior Member Evales's Avatar
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    Okay thanks guys.

    Just wondering where how the function comes to be divided by two.

    I've gotten to
    (sin x)^2 = 1 - cos(2x)
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  9. #9
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by Evales View Post
    Okay thanks guys.

    Just wondering where how the function comes to be divided by two.

    I've gotten to
    (sin x)^2 = 1 - cos(2x)
    \cos(2x)=1-2\sin^2(x)\Rightarrow{2\sin^2(x)=1-\cos(2x)}\Rightarrow{\sin^2(x)=\frac{1-\cos(2x)}{2}}
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  10. #10
    Junior Member Evales's Avatar
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    Oh lol Duh.

    Thanks for holding my hand through this process!
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  11. #11
    Math Engineering Student
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    \int{\sin ^{2}x\,dx}=\int{\sin x(-\cos x)'\,dx}=-\sin x\cos x+\int{\cos ^{2}x\,dx}.

    Now use \cos^2x=1-\sin^2x and the conclusion follows.

    (A related thread.)
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