# How do you evaluate $\int e^{-7x}\cos^3(4x)\sin^2(4x)dx$?

How do you evaluate the following integral

$$\int e^{-7x}\cos^3{4x}\sin^2{4x}\,dx ?$$

-
I use wolfram alpha. –  Quinn Culver Apr 30 '13 at 22:20
Extract one factor of $\cos 4x$ to attach to the differential, to serve with the substitution $u = \sin 4x$. Write the remaining factor $\cos^2 4x$ as $(1 - \sin^2 4x)$ using the Pythagorean Identity. Then multiply out the terms and split the integrand into two terms that may be integrated by parts. –  RecklessReckoner Apr 30 '13 at 22:22

If you know complex analysis, the following might be helpful:

\begin{align} \cos a &= \frac{ e^{ia} + e^{-ia} }{2} \\ \sin a &= \frac{ e^{ia} - e^{-ia} }{2i} \end{align}

-
Writing the integrand in terms of a complex exponential is nice, but not a method generally included in second-semester calculus. More's the pity... –  RecklessReckoner Apr 30 '13 at 22:22

Well, that was a little reckless, even for me... It doesn't work literally the way I suggested in my comment (forgot the substitution would have to be used in the exponential factor as well), but I believe you can do something like this:

Choose $u = e^{-7x}$ and $dv = \cos^3(4x) \sin^2(4x) dx$, so we have

$$du \ = \ -\frac{1}{7} \cdot e^{-7x} \ dx \ \text{and} \ v \ = \ \frac{1}{4} \int \ (1 \ - \ w^2 ) \cdot \ w^2 \ dw \ = \ \frac{1}{4} \int \ w^2 \ - \ w^4 \ dw \ ,$$

with $w = \sin 4x \ , \ dw = 4 \cos 4x \ dx$. Now, the $\int v \ du$ integral can be split into two separate integrations-by-parts, $\int \ e^{-7x} \sin^3(4x) \ dx$ and $\int \ e^{-7x} \sin^5(4x) \ dx$ , with the various multiplicative constants to be pulled together here and there...

-