Question

Integral cos(3x)^3sen(3x)^7dx

Answer 1

One way to do this is to exploit the Pythagorean identity,

`\cos^2x+\sin^2x=1`

to rewrite

`\cos^3(3x)=\cos(3x)\cos^2(3x)=\cos(3x)(1-\sin^2(3x))`

so that

`\displaystyle\int\cos^3(3x)\sin^7(3x)\,\mathrm dx=\int\cos(3x)\left(\sin^7(3x)-\sin^9(3x)\right)\,\mathrm dx`

Then substitute
`u=\sin(3x)` and
`\frac{\mathrm du}3=\cos(3x)\,\mathrm dx` to get the integral

`\displaystyle\frac13\int u^7-u^9\,\mathrm du=\frac13\left(\frac{u^8}8-(u^(10))/(10)\right)+C`

`=\boxed{(\sin^8(3x))/(24)-(\sin^(10)(3x))/(30)+C}`

which is one correct form of the antiderivative. There's no reason we can't use the identity from before to express the integrand in terms of powers of cos(3x) instead.