<s>[INST] Find the indefinite integral of...? I have spent well over an hour trying to solve this equation:
$$\int\cos^4{x}\sin^3{x}\, dx$$
I have tried substituting u as $\cos{x}$, $\sin{x}$, $\cos^4{x}$, and $sin^3{x}$ to no avail. How can I solve this? 
 [/INST] Through "brute force", you can just write out the values...there are several identities you need to get there:
$$
\sin(a)\sin(b) = \frac{\cos(a - b)-\cos(a + b)}{2} = \frac{\cos(b - a)-\cos(a + b)}{2} \\
\cos(a)\cos(b) = \frac{\cos(a - b)+\cos(a + b)}{2} = \frac{\cos(b - a) + \cos(a + b)}{2} \\
\sin(a)\cos(b) = \frac{\sin(a + b) + \sin(a - b)}{2}
$$
You can verify all of those properties from the angle addition properties:
\begin{align}
\cos(a + b) =&& \cos(a)\cos(b) - \sin(a)\sin(b) \\ 
\cos(a - b) =& \cos(a)\cos(-b) - \sin(a)\sin(-b)= &\cos(a)\cos(b) + \sin(a)\sin(b)\\
\sin(a +b) =&& \sin(a)\cos(b)  + \sin(b)\cos(a) \\
\sin(a - b) =& \sin(a)\cos(-b) + \sin(-b)\cos(a) =&\sin(a)\cos(b) - \sin(b)\cos(a)
\end{align}
This gives:
\begin{align}
\cos^4(x)sin^3(x) = &\cos(x)\left(\sin(x)\cos(x)\right)^3 \\
=& \cos(x)\frac{\sin^3(2x)}{2^3} = \cos(x)\sin(2x)\sin^2(2x) \\
=& \frac{1}{8}\cos(x)\sin(2x)\frac{1 - \cos(4x)}{2} \\
\cos(x)\sin(2x) =& \frac{\sin(3x) + \sin(x)}{2} \\
\cos^4(x)\sin^3(x) = & \left.\left.\frac{1}{32}\right(\sin(3x) - \sin(3x)\cos(4x) + \sin(x) - \sin(x)\cos(4x)\right) \\
=& \left.\left.\frac{1}{32}\right(\sin(3x) - \sin(x)\right)  - \left.\left.\frac{1}{64} \right(\sin(7x) + \sin(x) + \sin(5x) + \sin(3x)\right) \\
=& \left.\left.\frac{1}{64}\right(-3\sin(x) + \sin(3x) - \sin(5x) - \sin(7x)\right)
\end{align}
This then gives the integral easily as:
$$
\int\cos^4(x)\sin^3(x)dx = \left.\left.\frac{1}{64}\right(3\cos(x) - \frac{1}{3}\cos(3x) + \frac{1}{5}\cos(5x) + \frac{1}{7}\cos(7x)\right) + C
$$
</s>