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  1. #1
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    tricky integral

    what do i do here?

    S 1/2sin^2(x) dx

    the S is meant to be the integrate zigzag thing so its integrate 1/2sin^2(x) dx? i am lost. i calculated the answer to be 1/2tan(x) using software but i don't get why so can someone talk me through a step by step?
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  2. #2
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    \displaystyle \sin^2 x = \frac{1-\cos 2x}{2}
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  3. #3
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    but it is 1 / (2sin^2(x)) so what do i get?

    1 / 2((1 - cos2x)/2)

    cancels to 1/1 - cos2x). i dont still know what to do when you integrate something in the bottom unless its exact derivative on top then you use logs but if it isn't... :s

    is it a substitution?

    thanks so far
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  4. #4
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    Yes, let u = 1-\cos(2x)
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  5. #5
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by mathcore View Post
    but it is 1 / (2sin^2(x)) so what do i get?

    1 / 2((1 - cos2x)/2)

    cancels to 1/1 - cos2x). i dont still know what to do when you integrate something in the bottom unless its exact derivative on top then you use logs but if it isn't... :s

    is it a substitution?

    thanks so far
    So the problem is actually
    \displaystyle \int \frac{1}{sin^2(x)}dx

    (Please make better use of parethesis!)

    You might recognize that
    \displaystyle \int \frac{1}{sin^2(x)}dx = \int csc^2(x) dx = -cot(x) + C

    If not, then
    \displaystyle \int \frac{1}{sin^2(x)}dx = \int \frac{sin^2(x) + cos^2(x)}{sin^2(x)}dx

    Now let y = cos(x) and see what happens.

    -Dan

    (This is probably the longer way around.)
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  6. #6
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    e^(i*pi)'s Avatar
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    Take a factor of 1/2 out the integral then continue as topsquark did above

    \displaystyle \int \dfrac{1}{2 \sin^2(x)} = \int \left( \dfrac{1}{2} \cdot \csc^2(x)\right) = \dfrac{1}{2} \int \csc^2(x)


    You also have the option of using the double angle identity and a sub as per posts 2 and 4


    edit: corrected "sec" to "csc"
    Last edited by e^(i*pi); February 21st 2011 at 04:01 PM.
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  7. #7
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    Quote Originally Posted by e^(i*pi) View Post
    Take a factor of 1/2 out the integral then continue as topsquark did above

    \displaystyle \int \dfrac{1}{2 \sin^2(x)} = \int \left( \dfrac{1}{2} \cdot \sec^2(x)\right) = \dfrac{1}{2} \int \sec^2(x)


    You also have the option of using the double angle identity and a sub as per posts 2 and 4
    if you mean csc instead of sec then the solution i have works... can you tell me is it possible you meant csc?

    my solution is -1/(2tan(x)).
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  8. #8
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    e^(i*pi)'s Avatar
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    Quote Originally Posted by mathcore View Post
    if you mean csc instead of sec then the solution i have works... can you tell me is it possible you meant csc?

    my solution is -1/(2tan(x)).
    Yes it is csc, such is what happens when doing calculus at 1am. Thanks for pointing it out.


    Your answer is almost correct - what do you need to add for indefinite integrals?
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  9. #9
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    no i dont need a constant in this case because its part of a seperating variables d.e., i have the constant on the other side
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  10. #10
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by mathcore View Post
    what do i do here?

    S 1/2sin^2(x) dx
    Quote Originally Posted by mathcore View Post
    no, the problem is NOT 1/sin^2(x), it is 1/(2sin^2(x)) which is what i Said. do NOT call into question my parenthesis when you havent got a clue. my parenthesis was correct.
    Funny how that 1/2sin^2(x) is \displaystyle \frac{1}{2} sin^2(x) and 1/(2sin^2(x)) is \displaystyle \frac{1}{2 sin^2(x)}. These are obviously not the same. Speaking of obvious
    \displaystyle \int \frac{1}{2 sin^2(x)} dx = \frac{1}{2} \int \frac{1}{sin^2(x)} dx

    I expected you to be able to recognize that I factored the 2 out and worked on the meat of the problem. Evidently I expected too much from you.

    -Dan
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  11. #11
    Rhymes with Orange Chris L T521's Avatar
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    The question has been answered. This is yet another good reason why members should learn the basics of LaTeX, so that no miscommunication occurs.

    Thread Closed.
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