Integration problem

**Jgirl689** · March 9th 2010, 07:11 PM

Evaluate the integral

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

I have no idea how to solve..

**TKHunny** · March 9th 2010, 07:15 PM

Integration by parts will produce a reduction formula.

**Prove It** · March 9th 2010, 07:21 PM

Originally Posted by Jgirl689

Evaluate the integral

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

I have no idea how to solve..

$\int{\cos^5{x}\,dx} = \int{\cos^4{x}\cos{x}\,dx}$

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

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

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

Now make the substitution $u = \sin{x}$ so that $\frac{du}{dx} = \cos{x}$ .

**Pulock2009** · March 9th 2010, 09:34 PM

Originally Posted by Jgirl689

Evaluate the integral

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

I have no idea how to solve..

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.

**Prove It** · March 9th 2010, 09:52 PM

Originally Posted by Pulock2009

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.

Another alternative is to use trigonometric identities to convert it into functions such as $\sin{nx}$ or $\cos{nx}$ .

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