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    For this problem: \int \frac{4}{sin^2x}\ dx

    I have gone about my working in this way:

    \int \frac{4}{sin^2x}\ dx = -1 \times 4 \times -cosx \times (sinx)^{-1} +C

    = \frac{4cosx}{sinx}\ + C


    I think I have done something wrong.... :s
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    if you integrate sec2 you get tan

    if you integrate cosec2 you get.....
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    (Original post by the bear)
    if you integrate sec2 you get tan

    if you integrate cosec2 you get.....
    Wait.. why is this to do with sec2 and cosec2's ?
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    (Original post by Funky_Giraffe)
    Wait.. why is this to do with sec2 and cosec2's ?
    because

    \frac{1}{sin^2x}\ is cosec2x
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    It's not exactly clear what you did in your workings but you appear to be missing a minus sign
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    (Original post by Funky_Giraffe)
    For this problem: \int \frac{4}{sin^2x}\ dx

    I have gone about my working in this way:

    \int \frac{4}{sin^2x}\ dx = -1 \times 4 \times -cosx \times (sinx)^{-1} +C

    = \frac{4cosx}{sinx}\ + C


    I think I have done something wrong.... :s
    Your workings are incorrect.

    Inspecting what you have written, I believe you were attempting to use the proven result of \int   [f(x)]^n  * f'(x) dx  =   \frac{ [f(x)]^{n+1}} {(n+1)}\ +C    , where f(x)= sin x and n= -2.

    Unfortunately, this is not feasible because f'(x)=cos x is not present in the original integral. For some unknown reason you made this term surface all of a sudden, which makes no sense.

    Like the rest have mentioned, there is a simple result for integrating cosec^2 x wrt x.

    Hope this helps. Peace.
 
 
 
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