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

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

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

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.