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

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- Dec 1st 2008, 11:34 AMcharlie8529Integration
What is the integral of: (4sin(x))/(2+cos(x)) and please explain the method. Thank you.

- Dec 1st 2008, 12:02 PMSoroban
Hello, charlie8529!

Quote:

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

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?

- Dec 1st 2008, 12:03 PMvincisonfire
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. - Dec 1st 2008, 12:15 PMcharlie8529
Thanks for the help guys.