1. ## 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

2. Originally Posted by BiGpO6790
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

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

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

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

4. Originally Posted by BiGpO6790
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
$\displaystyle \int {{{\tan }^5}xdx} = \int {\frac{{{{\sin }^5}x}} {{{{\cos }^5}x}}dx} = \int {\frac{{\sin x{{\left( {1 - {{\cos }^2}x} \right)}^2}}} {{{{\cos }^5}x}}dx} = \left\{ \begin{gathered} \cos x = u, \hfill \\ \sin xdx = - du \hfill \\ \end{gathered} \right\} =$

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

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

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

5. Originally Posted by BiGpO6790
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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# integral tan 5x dx

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