1. ## Integration

What is the integral of: (4sin(x))/(2+cos(x)) and please explain the method. Thank you.

2. Hello, charlie8529!

$\int \frac{4\sin x}{2+\cos x}\,dx$

We have: . $4 \int\frac{1}{2+\cos x}\,(\sin x\,dx)$

Let: $u \:=\:2 + \cos x\quad\Rightarrow\quad du \:=\:-\sin x\,dx \quad\Rightarrow\quad \sin x\,dx \:=\:-du$

Substitute: . $4\int\frac{1}{u}\,(-du) \;=\;-4\int\frac{du}{u}$

Got it?

3. You want to compute the antiderivative of
$\frac{4sin(x)}{2+cos(x)}$
That is $\int \frac{4sin(x)}{2+cos(x)}dx$.
Let $u = 2 + cos(x)$
Then you can differentiate both side to find that
$du = -sin(x)dx$
We notice that in the integral above we have $4 sin(x)dx$ at the denominator.
This corresponds to $-4du = -4(-sin(x))dx = 4 sin(x) dx$
We can now substitute in the integral to simplify the notation.
$\int \frac{4sin(x)}{2+cos(x)}dx = \int \frac{-4du}{u}dx = -4 \int \frac{du}{u}dx$
This should now remember you something.
$-4 \int \frac{du}{u}dx = -4 ln|u| = -4 ln|2+cos(x)|$ because we substitute math] u = 2 + cos(x) [/tex].
$ln|\frac{1}{(2+cos(x))^4}|+C$ where C is a constant number. Don't forget it. I use to and teachers will take off mark any time for this little mistake.