# Integration of Cosine

1. Jan 8, 2013

### golanor

1. The problem statement, all variables and given/known data

$\int { { cos }^{ 2n+1 }(x)dx }$

2. Relevant equations
${ cos }^{ 2 }+{ sin }^{ 2 } = 1$

3. The attempt at a solution
i got to here:
$\int { { (1-{ sin }^{ 2 }(x)) }^{ n }d(sin(x)) }$

Any help would be appreciated!

2. Jan 8, 2013

### micromass

Staff Emeritus
So you got to find

$$\int (1-x^2)^ndx.$$

Use the binomial theorem and linearity of the integral.

3. Jan 8, 2013

### Ray Vickson

There is no simple formula for the result, unless you want to express it in terms of a hypergeometric function. The answer is a polynomial of degree (2n+1) in $\sin(x)$.

BTW: in [t e x] you should write "\sin(x)" instead of "sin(x)": the difference is $\sin(x)$ (nice) vs. $sin(x)$ (not nice).

4. Jan 8, 2013

### golanor

Thanks for the tex hint..still getting the hang of it (used to microsoft equation editor =\)

Thanks, i actually read about it, thought that i should use it but decided not to.

In case anyone wants the answer, it is:
$$\int { { (1-{ \sin }^{ 2 }(x)) }^{ n }d(\sin (x)) } =\sin (x)-\frac { n{ \sin }^{ 3 }(x) }{ 3 } +...+\frac { { (-1) }^{ n }{ \sin }^{ 2n+1 }(x) }{ 2n+1 } +C$$