Trigonometric integral

1. Sep 25, 2005

Stevecgz

Problem:
$$\int sin^6 x dx$$
Progress so far:
$$\int (sin^2 x)^3 dx$$
$$\frac{1}{8} \int (1-cos2x)^3 dx$$
$$\frac 1 8 \int (1 - 3cos2x + 3cos^22x - cos^32x) dx$$

Any help is appreciated.

I can see using a half angle identity for cos^2(2x), but what do I do with the cos^3(2x)?

Steve

Last edited: Sep 25, 2005
2. Sep 25, 2005

whozum

Try looking up some sine integral reduction formulas on google. They take care of integrals involving powers of sine pretty nicely.

3. Sep 25, 2005

Stevecgz

I've found one in my text. Would I simply continue using the reduction formula until I get to sin^0(x)?

Steve

4. Sep 25, 2005

whozum

Yes thats pretty much how we did it.

5. Sep 25, 2005

Stevecgz

Thanks whozum.

Steve