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Math Help - 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

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

    = - \int {\frac{{{{\left( {1 - {u^2}} \right)}^2}}}<br />
{{{u^5}}}du}  = \int {\frac{{ - 1 + 2{u^2} - {u^4}}}<br />
{{{u^5}}}du}  = \int {\left( { - \frac{1}<br />
{{{u^5}}} + \frac{2}<br />
{{{u^3}}} - \frac{1}<br />
{u}} \right)du}  =  \ldots
    Last edited by DeMath; September 25th 2009 at 02:38 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
    \int {{{\tan }^5}xdx}  = \int {\frac{{{{\sin }^5}x}}<br />
{{{{\cos }^5}x}}dx}  = \int {\frac{{\sin x{{\left( {1 - {{\cos }^2}x} \right)}^2}}}<br />
{{{{\cos }^5}x}}dx}  = \left\{ \begin{gathered}<br />
  \cos x = u, \hfill \\<br />
  \sin xdx =  - du \hfill \\ <br />
\end{gathered}  \right\} =

    =  - \int {\frac{{{{\left( {1 - {u^2}} \right)}^2}}}<br />
{{{u^5}}}du}  = \int {\frac{{ - 1 + 2{u^2} - {u^4}}}<br />
{{{u^5}}}du}  = \int {\left( { - \frac{1}<br />
{{{u^5}}} + \frac{2}<br />
{{{u^3}}} - \frac{1}<br />
{u}} \right)du}  =

    = \frac{1}<br />
{{4{u^4}}} - \frac{1}<br />
{{{u^2}}} - \ln \left| u \right| + C = \frac{1}<br />
{{4{{\cos }^4}x}} - \frac{1}<br />
{{{{\cos }^2}x}} - \ln \left| {\cos x} \right| + C =

    = \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:

    \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 u = \tan x. For the second integral:

    \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 u = \tan x. For the second integral:

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