# Trigonometric integral

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:

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

whozum said:
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

Yes thats pretty much how we did it.

Thanks whozum.

Steve