# What is the integral of 3sin(2x)?

Dec 16, 2014

$\int 3 \sin \left(2 x\right) \mathrm{dx} = - \frac{3 \cos \left(2 x\right)}{2} + C$

Explanation
You could solve this integral using $u$ -substitution. Typically, we pick $u$ to be our "inner" function, so here, let's denote $u = 2 x$ which is inside the $3 \sin \left(2 x\right) = 3 \sin \left(u\right)$. Now, we can integrate, but we need to make sure all of our variables are in terms of $u$, so we need to find $\mathrm{dx}$ in terms of $u$.

$u = 2 x$
$\to$$\mathrm{du} = 2 \mathrm{dx}$
$\to$ $\mathrm{dx} = \frac{1}{2} \mathrm{du}$

$\int 3 \sin \left(2 x\right) \mathrm{dx} = \int 3 \sin \left(u\right) \cdot \frac{1}{2} \mathrm{du} = \int \frac{3}{2} \sin \left(u\right) \mathrm{du} = - \frac{3}{2} \cos \left(u\right) + C$

Since the original question was in terms of $x$, we want to make sure our final answer is also in terms of $x$. Remember, $u = 2 x$ so we just sub back in:

$- \frac{3}{2} \cos \left(u\right) + C = - \frac{3}{2} \cos \left(2 x\right) + C$ Which you can rewrite as $- \frac{3 \cos \left(2 x\right)}{2} + C$ if you wanted to.

Alternatively...
You could also solve this integral through inspection, since we easily know the integral of $\sin \left(x\right)$ alone is supposed to be $- \cos \left(x\right) + C$. But we need the integral of $3 \sin \left(2 x\right)$. So we can pull out the constant 3 from the integral
$\int 3 \sin \left(2 x\right) \mathrm{dx} = 3 \int \sin \left(2 x\right) \mathrm{dx}$
Our lives would be easier if we had $2 \sin \left(2 x\right)$ because that is easily integrated...and we can do that! As long as we multiply by 1/2 on the outside of the integral to cancel it out. So we end up with:
$\int 3 \sin \left(2 x\right) \mathrm{dx} = 3 \int \sin \left(2 x\right) \mathrm{dx} = 3 \cdot \frac{1}{2} \int 2 \sin \left(2 x\right) \mathrm{dx} = - \frac{3}{2} \cos \left(2 x\right) + C$