Solution for the integral $∫ \sin(4x)(1+\cos(4x))\mathrm dx$

Question

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.

Answer 1

Method 1 -

Let

\begin{align} 1+\cos 4x &= u\\ -\sin 4x \cdot 4\ dx &= du\\ \sin 4x\ dx &= \frac 1{-4}\ du \end{align}

Putting these values,

$$\int \frac 1{-4} \cdot u\ du\\ \frac 1{-4} \int u\ du\\ \frac 1{-4} \cdot \frac {u^2}2 + c\\ \frac {u^2}{-8} + c$$

Put value of $u$ back.

$\frac {(1+\cos 4x)^2}{-8} + c$

Method 2 -

As you already done,

Multiply and divide $\sin4x \cos4x$ with 2.

We have,

$$\frac{2\sin4x\cos4x}{2}$$

$$\frac{\sin8x}{2}$$

As, $2 \sin4x \cos4x = \sin8x$.

You can then integrate easily.

Answer 2

I would write $$\int\sin(4x)dx$$ and $$\int \sin(4x)\cos(4x)dx$$ not that $$\sin(8x)=2\sin(4x)\cos(4x)$$ is hold

Answer 3

Let $u =1+ \cos 4x$, then $du = -4 \sin 4x$ $\implies -\frac{1}{4}du = \sin4x$

$$\int \sin4x(1+\cos4x)dx = -\frac{1}{4} \int u du = -\frac{1}{4}\frac{u^2}{2} + C = -\frac{1}{8}(1+\cos4x)^2 + C$$