# Integral of cos^5x dx?

Jun 7, 2018

$\frac{1}{80} \sin \left(5 x\right) + \frac{5}{48} \sin \left(3 x\right) + \frac{5}{8} \sin \left(x\right) + C$

#### Explanation:

$\int {\cos}^{5} \left(x\right) \mathrm{dx}$

If you know the complex exponential form of $\cos \left(x\right)$ then you could proceed using the following method:

${\cos}^{5} \left(x\right) = {\left(\frac{{e}^{i x} + {e}^{- i x}}{2}\right)}^{5}$

Now, expand this with the binomial theorem to get:

$= \frac{1}{32} \left({e}^{i 5 x} + 5 {e}^{i 3 x} + 10 {e}^{i x} + 10 {e}^{- i x} + 5 {e}^{- 3 i x} + {e}^{- 5 i x}\right)$

Now collect the terms like so:

$= \frac{1}{32} \left(\left\{{e}^{i 5 x} + {e}^{- i 5 x}\right\} + 5 \left\{{e}^{i 3 x} + {e}^{- i 3 x}\right\} + 10 \left\{{e}^{- i x} + {e}^{i x}\right\}\right)$

$= \frac{1}{16} \left(\frac{{e}^{i 5 x} + {e}^{- i 5 x}}{2} + 5 \frac{{e}^{i 3 x} + {e}^{- i 3 x}}{2} + 10 \frac{{e}^{- i x} + {e}^{i x}}{2}\right)$

$= \frac{1}{16} \cos \left(5 x\right) + \frac{5}{16} \cos \left(3 x\right) + \frac{5}{8} \cos \left(x\right)$

So from this it may follow that:

$\int {\cos}^{5} \left(x\right) \mathrm{dx} = \int \frac{1}{16} \cos \left(5 x\right) + \frac{5}{16} \cos \left(3 x\right) + \frac{5}{8} \cos \left(x\right) \mathrm{dx}$

Now integrate to get:

$\frac{1}{80} \sin \left(5 x\right) + \frac{5}{48} \sin \left(3 x\right) + \frac{5}{8} \sin \left(x\right) + C$