Solving an Integral of Trig Functions

**poXGxxi** · Sep 29th 2009, 04:34 PM

[IMG]file:///C:/Users/Beth/Desktop/integral.jpg[/IMG]

I have tried to solve this integral using several different trig identities, like the double angle formula, etc. When I use the identities, I am not getting a less complex integral, so I must not be using the right identity. Can someone please suggest a trig identity to use. Thank you.

**DeMath** · Sep 29th 2009, 04:58 PM

Originally Posted by poXGxxi

I have tried to solve this integral using several different trig identities, like the double angle formula, etc. When I use the identities, I am not getting a less complex integral, so I must not be using the right identity. Can someone please suggest a trig identity to use. Thank you.

Hint:

$\int {{{\sin }^5}x\,{{\cos }^4}x\,dx} = \int {\sin x\,{{\sin }^4}x\,{{\cos }^4}x\,dx} =$

$= \int {\sin x{{\left( {1 - {{\cos }^2}x} \right)}^2}{{\cos }^4}x\,dx} = \left\{ \begin{gathered}<br /> \cos x = u, \hfill \\<br /> \sin x\,dx = - du \hfill \\ <br /> \end{gathered} \right\} =$

$= - \int {{{\left( {1 - {u^2}} \right)}^2}\,{u^4}\,du} = \ldots$

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