Evaluate the integral
integral (0, pi/2) 60cos^5 xdx...
I have no idea how to solve..
$\displaystyle \int{\cos^5{x}\,dx} = \int{\cos^4{x}\cos{x}\,dx}$
$\displaystyle = \int{(\cos^2{x})^2\cos{x}\,dx}$
$\displaystyle = \int{(1 - \sin^2{x})^2\cos{x}\,dx}$
$\displaystyle = \int{(1 - 2\sin^2{x} + \sin^4{x})\cos{x}\,dx}$.
Now make the substitution $\displaystyle u = \sin{x}$ so that $\displaystyle \frac{du}{dx} = \cos{x}$.
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.