# Math Help - trigonometric integrals

1. ## trigonometric integrals

Hi the problem is to evaluate the following

integral sign ("S") b=pi/3 a=0 sin^3 x dx

so using the identity 1= sin^2 x + cos^2 x

S sin^3 x dx = S sin x (1-cos^2 x) dx

So if I let u= cos x du = sin x dx

Then the original equation becomes = S (1-u^2) du
= u - (1/3) u^3
Substituting back and now applying a= 0 and b = pi/3
= (cos pi/3 - (1/3) cos^3 pi/3) - (cos 0 - (1/3) cos^3 0)

First am I doing this right?? And second if this is right does cos^3 pi/3 = 1/8?

Thanks
Calculus Beginner

2. Originally Posted by calcbeg
Hi the problem is to evaluate the following

integral sign ("S") b=pi/3 a=0 sin^3 x dx

so using the identity 1= sin^2 x + cos^2 x

S sin^3 x dx = S sin x (1-cos^2 x) dx

So if I let u= cos x du = sin x dx ... du = -sin x dx

...

3. ## now regarding cos^3 pi/3

is cos^3 pi/3 = (1/2)^3 = 1/8???

Thanks

Calculus Beginner

4. Originally Posted by calcbeg
is cos^3 pi/3 = (1/2)^3 = 1/8???

Thanks

Calculus Beginner
yes