How Do You Integrate sin^6(x) Using Trigonometric Identities?

[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

 

[whozum]

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

 

[Stevecgz]

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

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

Steve

 

[whozum]

Yes that's pretty much how we did it.

 

[Stevecgz]

Thanks whozum.

Steve