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Math Help - Integration Help on trigo

  1. #1
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    Integration Help on trigo

    \int cos^4 2x sin^3 2x  dx
    I've tried simplifying the equation first but got stucked instead, and tried substitution also.
    Which method should i use for this?
    Thanks in advance.
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  2. #2
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    Re: Integration Help on trigo

    hmmm why don't you try to first factor sin^3(2x) to sin^2(2x)(sin(2x))... then since sin^2(2x) is also equal to 1-cos^2(2x).... you rearrange it to a indefinite integral of cos^4(2x)[1-cos^2(2x)]sin(2x)dx.. then do substitution... u=cos (2x).... du= (2)(-sin(2x))dx.... now... please do the rest
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  3. #3
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    Re: Integration Help on trigo

    Try rewriting the integrand as follows:

    \cos^4(2x)\sin^3(2x)=\cos^4(2x)\sin^2(2x)\sin(2x)=  \cos^4(2x)(1-\cos^2(2x))\sin(2x)=

    \cos^4(2x)\sin(2x)-\cos^6(2x)\sin(2x)

    Can you proceed from here using an appropriate substitution?
    Thanks from tempq1
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  4. #4
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    Re: Integration Help on trigo

    At my university, students are taught general methods for the integrals \int\cos^nx\sin^mx\ dx and \int\sec^nx\tan^mx\ dx. The substitution depends on whether n and m are even or odd - it's actually pretty straightforward if you look at the derivatives of those four trigonometric functions.

    - Hollywood
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  5. #5
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    Re: Integration Help on trigo

    Thanks guys question solved. Just another question
    \int cot^2 \pi x dx
    i tried changing it into
    \frac{1 + cos2\pi x}{1-cos2\pi x} and got stuck.
    ps:can't seem to edit my original post.
    Last edited by tempq1; December 14th 2012 at 07:34 PM.
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  6. #6
    MHF Contributor MarkFL's Avatar
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    Re: Integration Help on trigo

    I would use a Pythagorean identity to rewrite the integrand as:

    \csc^2(\pi x)-1

    and then use the fact that:

    \frac{d}{dx}(\cot(u(x)))=-\csc^2(u(x))\frac{du}{dx}

    to rewrite the integral as:

    -\frac{1}{\pi}\int\,d(\cot(\pi x))-\int\,dx
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