Definite Integral of $\sin^4(x)\cos(x).$ Evaluate $$\int_0^\pi \sin^4\theta \cos\theta\, d\theta.$$
This is clearly very simple under indefinite circumstances, that is, to substitute $u=\sin\theta$, and work from there, but that leaves me with $0-0$, which is wrong, certainly.
 A: It is not wrong. The indefinite integral evaluates to 
$$\sin^5(\theta)/5$$
so when you take $\sin^5(\pi)/5 - \sin^5(0)/5$ you get $0-0$ like you said. 
A: You don't need substitution you can do 
$$
\frac{d}{d\theta}\sin^5\theta = 5 \sin^4\theta \cos \theta
$$
This your integral is
$$
\int \frac{1}{5}\frac{d}{d\theta}\sin^5\theta d\theta
$$
Which means you just apply the boundary.
A: No need to substitute $u=\sin\theta$.
$$\int_0^{\pi}\sin^4\theta\cos\theta\,d\theta=\int_0^{\pi/2}\sin^4\theta\cos\theta\,d\theta+\int_0^{\pi/2}\sin^4(\pi-\theta)\cos(\pi-\theta)\,d\theta$$
$$=\int_0^{\pi/2}\sin^4\theta\cos\theta\,d\theta-\int_0^{\pi/2}\sin^4\theta\cos\theta\,d\theta=0$$
A: i don't think you even need to evaluate this integral. just the symmetry of $\sin$ and $\cos$ would do. that is we will use the facts $$\sin(\pi - t) = \sin t, \cos(\pi - t) = -\cos t.$$
we will make a change of variable  $$u =  \pi - \theta, \theta = \pi - u, du = -d \theta$$ so that $$I = \int_0^\pi \sin^4 \theta \, \cos \theta \, d \theta = \int_\pi^0\sin^4 u(-\cos u)(-d u)= -\int_0^\pi\sin^4 u \,cos u \,du =-I$$
so that $$2I = 0 \to I = 0. $$
A: If you are familiar with linear algebra, you can use the inner product to evaluate the integral. Since the $\sin^4(\theta)\cos(\theta)$ is even, 
$$
\int_0^{\pi}\sin^4(\theta)\cos(\theta)d\theta = 
\frac{1}{2}\int_0^{2\pi}\sin^4(\theta)\cos(\theta)d\theta\tag{1}
$$
The orthonormal basis for the functions $\sin(\theta)$ and $\cos(\theta)$ are
$$
\bigl\{1/\sqrt{2},\cos(n\theta),\sin(n\theta)\bigr\}
$$
where $n\in\mathbb{Z}$. The inner product of this space is
$$
\langle f,g\rangle = \frac{1}{\pi}\int_0^{2\pi}fg \ d\theta=\delta(f,g)=
\begin{cases}
1, & f=g\\
0, & f\neq g
\end{cases}
$$
We can write equation $(1)$ as
\begin{align}
\frac{1}{2}\int_0^{2\pi}\sin^4(\theta)\cos(\theta)d\theta & = \frac{\pi}{2}\biggl[\frac{1}{\pi}\int_0^{2\pi}\sin^4(\theta)\cos(\theta)d\theta\biggr]\\
&= \frac{\pi}{2}\biggl[\frac{1}{\pi}\int_0^{2\pi}\frac{1}{2\sqrt{2}}\frac{1}{\sqrt{2}}\cos(\theta)d\theta - \frac{1}{\pi}\int_0^{2\pi}\frac{1}{2}\cos(\theta)\cos(2\theta)d\theta\\
&+\frac{1}{\pi}\int_0^{2\pi}\frac{1}{4\sqrt{2}}\frac{1}{\sqrt{2}}\cos(\theta)d\theta + \frac{1}{\pi}\int_0^{2\pi}\frac{1}{8}\cos(\theta)\cos(4\theta)d\theta\biggr]\\
& = \frac{\pi}{4\sqrt{2}}\langle 1/\sqrt{2},\cos(\theta)\rangle - \frac{\pi}{4}\langle\cos(\theta),\cos(2\theta)\rangle + \frac{\pi}{8\sqrt{2}}\langle 1/\sqrt{2},\cos(\theta)\rangle\\
& + \frac{\pi}{16}\langle\cos(\theta),\cos(4\theta)\rangle
\end{align}
Since every inner product consists of $f\neq g$, every inner product is a constant times $\langle \ , \ \rangle = 0$; therefore, the integral is zero.
