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Thread: Integral of tan^5(x)dx

  1. #1
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    Integral of tan^5(x)dx

    I can't seem to find the integral of tan^5(x)
    I know that i have to use identities for this, but I've been looking at this problem for a while now and cant see where to go with it
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  2. #2
    Senior Member DeMath's Avatar
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    Quote Originally Posted by BiGpO6790 View Post
    I can't seem to find the integral of tan^5(x)
    I know that i have to use identities for this, but I've been looking at this problem for a while now and cant see where to go with it
    Hint

    $\displaystyle \int {{{\tan }^5}x\,dx} = \int {\frac{{{{\sin }^5}x}}
    {{{{\cos }^5}x}}\,dx} = \int {\frac{{\sin x{{\left( {1 - {{\cos }^2}x} \right)}^2}}}
    {{{{\cos }^5}x}}\,dx} = \left\{ \begin{gathered}
    \cos x = u, \hfill \\
    \sin xdx = - du \hfill \\
    \end{gathered} \right\} =$

    $\displaystyle = - \int {\frac{{{{\left( {1 - {u^2}} \right)}^2}}}
    {{{u^5}}}du} = \int {\frac{{ - 1 + 2{u^2} - {u^4}}}
    {{{u^5}}}du} = \int {\left( { - \frac{1}
    {{{u^5}}} + \frac{2}
    {{{u^3}}} - \frac{1}
    {u}} \right)du} = \ldots$
    Last edited by DeMath; Sep 25th 2009 at 01:38 PM. Reason: typo
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  3. #3
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    answer check

    Ok, i used ur advice and i went through the integration, as a result i got:

    -1/(4(cos^4(x)))+(1/(cos^2(x)))+ln(cos(x))+c

    i think that can be simplified down but is that the basis of a correct answer
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  4. #4
    Senior Member DeMath's Avatar
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    Quote Originally Posted by BiGpO6790 View Post
    Ok, i used ur advice and i went through the integration, as a result i got:

    -1/(4(cos^4(x)))+(1/(cos^2(x)))+ln(cos(x))+c

    i think that can be simplified down but is that the basis of a correct answer
    $\displaystyle \int {{{\tan }^5}xdx} = \int {\frac{{{{\sin }^5}x}}
    {{{{\cos }^5}x}}dx} = \int {\frac{{\sin x{{\left( {1 - {{\cos }^2}x} \right)}^2}}}
    {{{{\cos }^5}x}}dx} = \left\{ \begin{gathered}
    \cos x = u, \hfill \\
    \sin xdx = - du \hfill \\
    \end{gathered} \right\} =$

    $\displaystyle = - \int {\frac{{{{\left( {1 - {u^2}} \right)}^2}}}
    {{{u^5}}}du} = \int {\frac{{ - 1 + 2{u^2} - {u^4}}}
    {{{u^5}}}du} = \int {\left( { - \frac{1}
    {{{u^5}}} + \frac{2}
    {{{u^3}}} - \frac{1}
    {u}} \right)du} =$

    $\displaystyle = \frac{1}
    {{4{u^4}}} - \frac{1}
    {{{u^2}}} - \ln \left| u \right| + C = \frac{1}
    {{4{{\cos }^4}x}} - \frac{1}
    {{{{\cos }^2}x}} - \ln \left| {\cos x} \right| + C =$

    $\displaystyle = \frac{1}{4}{\sec ^4}x - {\sec ^2}x - \ln \left| {\cos x} \right| + C.$
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  5. #5
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    Quote Originally Posted by BiGpO6790 View Post
    I can't seem to find the integral of tan^5(x)
    I know that i have to use identities for this, but I've been looking at this problem for a while now and cant see where to go with it
    Alternatively:

    $\displaystyle \int \tan^5 x \, dx = \int (\sec^2 x - 1) \tan^3 x \, dx = \int \sec^2 x \tan^3 x \, dx - \int \tan^3 x \, dx$.

    For the first integral make the substitution $\displaystyle u = \tan x$. For the second integral:

    $\displaystyle \int \tan^3 x \, dx = \int (\sec^2 x - 1) \tan x \, dx = \int \sec^2 x \tan x \, dx - \int \tan x \, dx$.

    For the first integral make the substitution $\displaystyle u = \tan x$. For the second integral:

    $\displaystyle \int \tan x \, dx = \int \frac{\sin x}{\cos x} \, dx$ and you make the substitution $\displaystyle u = \cos x$.
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