Evaluate the integral

integral (0, pi/2) 60cos^5 xdx...

I have no idea how to solve..

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- Mar 9th 2010, 07:11 PMJgirl689Integration problem
Evaluate the integral

integral (0, pi/2) 60cos^5 xdx...

I have no idea how to solve.. - Mar 9th 2010, 07:15 PMTKHunny
Integration by parts will produce a reduction formula.

- Mar 9th 2010, 07:21 PMProve It
$\displaystyle \int{\cos^5{x}\,dx} = \int{\cos^4{x}\cos{x}\,dx}$

$\displaystyle = \int{(\cos^2{x})^2\cos{x}\,dx}$

$\displaystyle = \int{(1 - \sin^2{x})^2\cos{x}\,dx}$

$\displaystyle = \int{(1 - 2\sin^2{x} + \sin^4{x})\cos{x}\,dx}$.

Now make the substitution $\displaystyle u = \sin{x}$ so that $\displaystyle \frac{du}{dx} = \cos{x}$. - Mar 9th 2010, 09:34 PMPulock2009many ways to solve this
i find several ways to solve this:

1)you can expand cos^5x in terms of multiples of x using euler's formula considering y=cosx+isinx and 1/y=cosx-isinx.

2)you can use a reduction formula as TKHunny said and

3)the way 'prove it' did it

4)and lastly u can perhaps use some property of definite integrals:

integration(0,pi/2)f(x)=integration(0,pi/2)f(pi/2-x).

i have not tested any of these methods though but i think they are all very feasible. - Mar 9th 2010, 09:52 PMProve It