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Math Help - Solve The Following Trigonometric Integral. (8)

  1. #1
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    Solve The Following Trigonometric Integral. (8)

    Evaluate

    \int \dfrac{cos^4(\theta)}{sin(\theta)} \, d\theta
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  2. #2
    Super Member girdav's Avatar
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    Since \int\frac{\cos^4 t}{\sin t}dt = \int\frac{\sin t\cos^4 t}{1-\cos^2t}dt, use the substitution x =\cos t.
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  3. #3
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    What I might do is

    \int \dfrac{\cos ^4 x}{\sin x}\,dx = \int \dfrac{\left(1 - \sin^2 x\right)^2}{\sin x}\,dx = \int \left(\csc x - 2 \sin x +  \sin^3x\right)\,dx.

    It might be a little easier.
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  4. #4
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    Quote Originally Posted by Danny View Post
    What I might do is

    \int \dfrac{\cos ^4 x}{\sin x}\,dx = \int \dfrac{\left(1 - \sin^2 x\right)^2}{\sin x}\,dx = \int \left(\csc x - 2 \sin x +  \sin^3x\right)\,dx.

    It might be a little easier.
    If you can remember off hand the antiderivatives of \displaystyle \csc{x} and \displaystyle \sin^3{x} (I know I can't ><, but that's just because I prefer to memorise procedures instead of integrals).
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  5. #5
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    Yes, but Danny's approach is for experienced people, since remaining integrals, at least, are easy.

    However girdav's solution is also good, and it only uses a substitution.
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  6. #6
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    Quote Originally Posted by Krizalid View Post
    Yes, but Danny's approach is for experienced people, since remaining integrals, at least, are easy.

    However girdav's solution is also good, and it only uses a substitution.
    Memorising integrals =/= Experience
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  7. #7
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    Nope, that's not my point.

    I was talking about practice, and it does have to do with intelligence.
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