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Math Help - Integral of (tanx)^3 * (secx)^88

  1. #1
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    Integral of (tanx)^3 * (secx)^88

    \int tan^3xsec^{88}x dx

    =\int tan^2xsec^{88}x tanx dx

    =\int (sex^2x-1)(sec^{88}x) tanx dx

    u=sec^2x, du=tanx

    =\int (u-1)u^{44} du

    =\int u^45-u^44 du

    =\frac{1}{46}u^{46}-\frac{1}{45}u^{45}

    =\frac{1}{46}(sec^2x)^{46}-\frac{1}{45}(sec^2x)^{45}

    =\frac{1}{46}sec^{94}x-\frac{1}{45}sec^{90}x

    Is this done correctly?
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  2. #2
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by circuscircus View Post
    u=sec^2x, du=tanx
    You've got it backward:
    \frac{d}{dx}tan(x) = sec^2(x)
    not the other way around.

    And then there's that comment about the sex...

    -Dan
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  3. #3
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    \int tan^3xsec^{88}x dx

    =\int tan^2xsec^{88}x tanx dx

    =\int (sec^2x-1)(sec^{87}x) secxtanx dx

    u=secx, du=secxtanx

    =\int (u^2-1)u^87 du

    =\int u^89-u^87 du

    =\frac{1}{90}u^90-\frac{1}{88}u^88

    =\frac{1}{90}sec^{90}x-\frac{1}{88}sec^{88}x

    So like this?
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  4. #4
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by circuscircus View Post
    \int tan^3xsec^{88}x dx

    =\int tan^2xsec^{88}x tanx dx

    =\int (sec^2x-1)(sec^{87}x) secxtanx dx

    u=secx, du=secxtanx

    =\int (u^2-1)u^87 du

    =\int u^89-u^87 du

    =\frac{1}{90}u^90-\frac{1}{88}u^88

    =\frac{1}{90}sec^{90}x-\frac{1}{88}sec^{88}x

    So like this?
    It looks right to me, except for some LaTeX coding errors.

    Oh yeah, and that pesky arbitrary constant...

    -Dan
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