How to solve this simple integral?
$\int_a^b\cos\theta \sin\theta \,d\theta$
I tried integrating by parts: given the rule
$\int u dv = uv - \int v du$
I substituted $\cos \theta$ as $dv $ and $\sin \theta$ as $u$ but I end up obtaining
$0 = \sin \theta \sin \theta$
You don't need integration by parts. Let $u=\sin\theta$ which means $du=\cos\theta d\theta$.
Integration by parts works:
$u' = \cos \phi, v = \sin \phi, u = \sin \phi , v' = \cos \phi$ then
$$ \int_a^b u' v = \int_a^b \cos \phi \sin \phi d \phi = (\sin \phi)^2 - \int_a^b \sin \phi \cos \phi d \phi $$
Therefore
$$ 2 \int_a^b \cos \phi \sin \phi d \phi = \sin^2 \phi$$
Using integration by parts, you should actually get: $$\int \cos \theta\sin\theta\,d\theta = \sin^2\theta - \int\cos \theta\sin\theta\,d\theta$$ If $A=B-A$, then $A=B/2$. Here, $A$ is your integral, and $B$ is $\sin^2 \theta$.
There should be a constant in there, too, just for kicks.
Using by parts, $\int_a^b\cos\theta\sin\theta d\theta=-\cos^2\theta|_a^b-\int_a^b\sin\theta\cos\theta d\theta\implies \int_a^b\cos\theta\sin\theta d\theta=\frac{1}{2}(\cos^2a-\cos^2b)$
For any derivable function $\,f(x)\,$ , we have
$$\int f'(x)f(x)^ndx=\frac{f(x)^{n+1}}{n+1}+C\;\;,\;\;n\neq -1\Longrightarrow$$
$$\int\limits_1^b\sin\theta\cos\theta \,d\theta=\int\limits_a^b\sin\theta (\sin\theta)'\,d\theta=\left.\frac{\sin^2\theta}{2}\right|_a^b=\ldots$$
