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Math Help - intergral of sin^4(x)cos^2(x)

  1. #1
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    intergral of sin^4(x)cos^2(x)

    hi i am running into trouble using trig subsitution in this particular problem so far the farthest that i have got to is

    1/8 times the intergral of 5/2 - 1/2cos(2x)+ 3/4cos(4x)+ 1/2 cos(4x)cos(2x)



    i really appreacated the help and thank you
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  2. #2
    MHF Contributor harish21's Avatar
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    Quote Originally Posted by momo55 View Post
    hi i am running into trouble using trig subsitution in this particular problem so far the farthest that i have got to is

    1/8 times the intergral of 5/2 - 1/2cos(2x)+ 3/4cos(4x)+ 1/2 cos(4x)cos(2x)



    i really appreacated the help and thank you

    How did you get to your result? Please try to show some work. Also, you will have to use the reduction formula for integration. Read this. It might help.
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  3. #3
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    I already know

    that you'll benefit from sin^2(x) = 1 - cos^2(x) along with integration by parts.
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  4. #4
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    \int\sin^{4}(x)\cos^{2}(x)dx

    \int\left(\sin(x)\cos(x)\right)^{2}\sin^{2}(x)dx

    \int\left(\frac{\sin(2x)}{2}\right)^{2}\left(\frac  {1-\cos(2x)}{2}\right)dx

    \frac{1}{8}\int\sin^{2}(2x)\left(1-\cos(2x)\right)dx

    \frac{1}{8}\int\left(\sin^{2}(2x)-\sin^{2}(2x)\cos(2x)\right)dx

    \frac{1}{8}\int\left[\left(\frac{1-\cos(4x)}{2}\right)-\sin^{2}(2x)\cos(2x)\right]dx

    \frac{1}{8}\int\left[\frac{1}{2}-\frac{\cos(4x)}{2}-\sin^{2}(2x)\cos(2x)\right]dx

    \frac{1}{8}\left[\frac{x}{2}-\frac{\sin(4x)}{8}-\frac{1}{6}\sin^{3}(2x)\right]
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