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Integration --Trig functions

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  1. Oct 5, 2015 #1
    1. The problem statement, all variables and given/known data
    integrate tan^3(9x)

    2. Relevant equations
    tan^2(x)= sec^2(x)-1
    integral of tanxdx= ln|secx|

    3. The attempt at a solution
    Integrate tan3(9x)dx
    Integral (sec2(x)-1)tan(9x)dx
    so then I distribute the tan(9x) giving me:
    integral tan(9x)sec2(x)dx - integral tan(9x) dx
    so then I solve the first part with u-substitution:
    u=tan(9x)
    (1/9)du=sec2(9x)
    and plug it in the first part
    (1/9)integral u - integral tan(9x)
    so solving for both
    (1/18)tan2(9x)-(1/9)ln |(sec(9x)| +c

    (I hope I typed it all correctly)

    Any help is appreciated!
     
  2. jcsd
  3. Oct 5, 2015 #2
    Just try something like you've already done, but only with the sin^3 in the numerator.

    Chet
     
  4. Oct 5, 2015 #3
    Wait so solve sin^3(9x)?
    Not sure how that would help but maybe it would go something like:

    integral sin3(9x)dx
    integral sin2(9x)sin(9x)
    we know that sin^2(9x)= 1-cos^2(9x)
    so change it up and distribute giving
    integral sin(9x)-integral sin(9x)cos2(9x)
    which gives me
    (-1/9)cos(9x)-integral sin(9x)cos2(9x)
    we use u-sub on the second part
    u=cos(9x)
    (1/9)du=-sin(9x)dx
    put it all in giving us
    -(1/9)cos(9x)+(1/27)sin^3(9x)+c

    Still kinda clueless on the original one though.
     
  5. Oct 5, 2015 #4
    $$\tan^3θ=\frac{\sin^3θ}{\cos^3θ}=\frac{(1-\cos^2θ)}{\cos^3θ}\sinθ$$
     
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