Thread: Solving an Integral of Trig Functions

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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.

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:

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

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

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