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.

  • 2
    $\begingroup$ Hint: the factor outside the parentheses is (essentially) the derivative of the function inside the parentheses. The chain rule is your friend here. $\endgroup$ – Ethan Bolker Mar 31 '17 at 18:12
  • $\begingroup$ sin8x = 2sin4xcos4x $\endgroup$ – novak Mar 31 '17 at 18:14

Method 1 -

Put $1+\cos 4x = u$

$-\sin 4x \cdot 4 dx = du$

$\sin 4x dx = \frac 1{-4} du$

Put 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,



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

You can then integrate easily.


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

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


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


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