Good day!
I'm kind of stuck in solving this integral problem. I've tried multiple times, but I'm not still not sure what is the correct way of solving.
The problem is to solve
$$\int \sin(4x)(1+\cos(4x)) \mathrm dx$$
I've tried two methods:
The first one is where I didn't distribute $\sin (4x)$, and the second is where I distributed $\sin (4x)$ to turn the function into
$\int \left(\sin(4x)+(\sin(4x)\cos(4x))\right) \mathrm dx$.
I'm not sure if I'm going to use trig. identities, or just straight up integrate by using $u = \cos (4x)$. Can someone help? Thanks in advance.
Best Answer
Let $u =1+ \cos 4x$, then $du = -4 \sin 4x$ $\implies -\frac{1}{4}du = \sin4x$
$$\int \sin4x(1+\cos4x)dx = -\frac{1}{4} \int u du = -\frac{1}{4}\frac{u^2}{2} + C = -\frac{1}{8}(1+\cos4x)^2 + C$$