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Math Help - trig integration

  1. #1
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    trig integration

    integrate 1/sinx(cosx)^4 dx
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  2. #2
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    Quote Originally Posted by twilightstr View Post
    integrate 1/sinx(cosx)^4 dx
    Substitute u = \cos x \Rightarrow dx = - \frac{du}{\sin x}. Note that \sin^2 x = 1 - \cos^2 x = 1 - u^2 \, ....
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  3. #3
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    i still dont know how to solve the problem.
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  4. #4
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    Quote Originally Posted by twilightstr View Post
    i still dont know how to solve the problem.
    Please show how you've tried to use the suggestions I've made. Do you know how to substitute? Where do you get stuck?
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  5. #5
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    yes i dont know how to substitute.
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  6. #6
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    Quote Originally Posted by twilightstr View Post
    yes i dont know how to substitute.
    Couldn't you get at least the first line: \int \frac{1}{\sin x} \, \frac{1}{u^4} \, \left( \frac{-du}{\sin x} \right).


    Then:

    = - \int \frac{1}{\sin^2 x} \, \frac{1}{u^4} \, du


    = - \int \frac{1}{1 - u^2} \, \frac{1}{u^4} \, du


     = - \int \frac{1}{(1 - u^2) u^4} \, du


    Now do a partial fraction decomposition.

    If you're doing this sort of question it's expected that you've seen the various techniques.
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  7. #7
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    Following up on mr fantastic's last integral we have,

    \frac{1}{\left( 1-{{u}^{2}} \right){{u}^{4}}}=\frac{{{u}^{4}}+\left( 1+{{u}^{2}} \right)\left( 1-{{u}^{2}} \right)}{\left( 1-{{u}^{2}} \right){{u}^{4}}}=\frac{1}{1-{{u}^{2}}}+\frac{1+{{u}^{2}}}{{{u}^{4}}}.

    Besides \frac{1}{1-{{u}^{2}}}=\frac{(1-u)+(1+u)}{2(1+u)(1-u)}=\frac{1}{2}\left( \frac{1}{1+u}+\frac{1}{1-u} \right) and \frac{1+{{u}^{2}}}{{{u}^{4}}}={{u}^{-4}}+{{u}^{-2}} so you got easy integrals to solve!
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