$$\int6\sin^3(5x)\cos^3(5x)\,\mathrm dx$$
Evaluate the integral above (include C as a constant)
I've gotten to the point in the question where I'm supposed to sub u into the equation and there's 1 part that I don't particularly understand. I've subbed cos(5x)=u and sin5x=(-1/2)du and came to the point: $$3\int (u^5-u^3)du$$ And apparently the next step is: $$3\int \frac{1}{6}u^5-\frac{1}{4}u^3+C$$ And I came up with $$3\int {6}u^5-{4}u^3+C$$ Is it because of something to do with the du component?
Do you mean $\int \sin^3(5x) \cos^3(5x)\mathrm dx$? Then, if you want to use the substitution $u=\cos(5x)$ then write the integrand as $$\sin^2(5x)\cos^3(5x) \sin(5x)=(1-\cos^2(5x))\cos^3(5x)\sin(x)=(\cos^3(5x)-\cos^5(5x))\sin(x)$$
Then, with $u=\cos(5x)$,
$\mathrm du= -5\sin(5x)\mathrm dx$
and the integral becomes
$-5\int (\cos^3(5x)-\cos^5(5x))\sin(5x)\mathrm dx$
$=-5\int\left(u^3- u^5\right)\mathrm du$
$=-5\left(\frac{1}{4}u^4- \frac{1}{6}u^6\right)+ C$
$= -\frac{5}{4}\cos^4(5x)+ \frac{5}{6}\cos^6(5x)+ C$