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
$\displaystyle \int {{{\tan }^5}x\,dx} = \int {\frac{{{{\sin }^5}x}}
{{{{\cos }^5}x}}\,dx} = \int {\frac{{\sin x{{\left( {1 - {{\cos }^2}x} \right)}^2}}}
{{{{\cos }^5}x}}\,dx} = \left\{ \begin{gathered}
\cos x = u, \hfill \\
\sin xdx = - du \hfill \\
\end{gathered} \right\} =$
$\displaystyle = - \int {\frac{{{{\left( {1 - {u^2}} \right)}^2}}}
{{{u^5}}}du} = \int {\frac{{ - 1 + 2{u^2} - {u^4}}}
{{{u^5}}}du} = \int {\left( { - \frac{1}
{{{u^5}}} + \frac{2}
{{{u^3}}} - \frac{1}
{u}} \right)du} = \ldots$
$\displaystyle \int {{{\tan }^5}xdx} = \int {\frac{{{{\sin }^5}x}}
{{{{\cos }^5}x}}dx} = \int {\frac{{\sin x{{\left( {1 - {{\cos }^2}x} \right)}^2}}}
{{{{\cos }^5}x}}dx} = \left\{ \begin{gathered}
\cos x = u, \hfill \\
\sin xdx = - du \hfill \\
\end{gathered} \right\} =$
$\displaystyle = - \int {\frac{{{{\left( {1 - {u^2}} \right)}^2}}}
{{{u^5}}}du} = \int {\frac{{ - 1 + 2{u^2} - {u^4}}}
{{{u^5}}}du} = \int {\left( { - \frac{1}
{{{u^5}}} + \frac{2}
{{{u^3}}} - \frac{1}
{u}} \right)du} =$
$\displaystyle = \frac{1}
{{4{u^4}}} - \frac{1}
{{{u^2}}} - \ln \left| u \right| + C = \frac{1}
{{4{{\cos }^4}x}} - \frac{1}
{{{{\cos }^2}x}} - \ln \left| {\cos x} \right| + C =$
$\displaystyle = \frac{1}{4}{\sec ^4}x - {\sec ^2}x - \ln \left| {\cos x} \right| + C.$
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 $\displaystyle u = \cos x$.