What is the integral of: (4sin(x))/(2+cos(x)) and please explain the method. Thank you.
Hello, charlie8529!
$\displaystyle \int \frac{4\sin x}{2+\cos x}\,dx$
We have: .$\displaystyle 4 \int\frac{1}{2+\cos x}\,(\sin x\,dx) $
Let: $\displaystyle u \:=\:2 + \cos x\quad\Rightarrow\quad du \:=\:-\sin x\,dx \quad\Rightarrow\quad \sin x\,dx \:=\:-du$
Substitute: .$\displaystyle 4\int\frac{1}{u}\,(-du) \;=\;-4\int\frac{du}{u}$
Got it?
You want to compute the antiderivative of
$\displaystyle \frac{4sin(x)}{2+cos(x)} $
That is $\displaystyle \int \frac{4sin(x)}{2+cos(x)}dx $.
Let $\displaystyle u = 2 + cos(x) $
Then you can differentiate both side to find that
$\displaystyle du = -sin(x)dx $
We notice that in the integral above we have $\displaystyle 4 sin(x)dx $ at the denominator.
This corresponds to $\displaystyle -4du = -4(-sin(x))dx = 4 sin(x) dx$
We can now substitute in the integral to simplify the notation.
$\displaystyle \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.
$\displaystyle -4 \int \frac{du}{u}dx = -4 ln|u| = -4 ln|2+cos(x)|$ because we substitute math] u = 2 + cos(x) [/tex].
You may use logarithm properties to express your answer as
$\displaystyle ln|\frac{1}{(2+cos(x))^4}|+C$ 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.