# Calculus help

by SEG9585
Tags: calculus
 P: 988 $$\int \sin^3{3x} \cos^3{3x}\,dx$$ First things first, get rid of the 3x's with a u substitution. It's easy to see that all it does is change the solution by a factor of 1 over 3. So we have, $$\frac{1}{3}\int \sin^3{u} \cos^3{u}\,du$$ If we had only 1 sine term or only 1 cosine term, we'd be gold. Problem solved. But we got 2 too many. So let's get rid of them! $$\sin^2{x} + \cos^2{x} = 1$$ Use this to turn the integral into $$\frac{1}{3}\int \sin^3{u} (1 - \sin^2{u}) \cos{u}\,du$$ which is easily separated and solved by substitution. Have another shot at the second one, keeping in mind that $$\frac{d}{dx}\tan{x} = \sec^2{x}$$ cookiemonster
 Sci Advisor HW Helper P: 2,537 Where did you get $$\cos^3=1-\sin^2$$ I would probably use the half angle formulas: $$\sin(2x)=2\sin(x)\cos(x)$$ so $$\sin(x)\cos(x)=\frac{\sin(2x)}{2}$$ so $$\sin(3x)\cos(3x)=\frac{\sin(6x)}{2}$$ Now you've got: $$\int\sin^3(3x)\cos^3(3x)dx$$ $$\int(\sin(3x)\cos(3x))^3dx$$ $$\frac{1}{8}\int\sin^3(6x)dx$$ Now $$\sin(3x)=3\sin(x)-4\sin^3(x)$$ so $$\sin^3(6x)=\frac{3\sin(6x)-\sin(18x)}{4}$$ so $$\frac{1}{32}\int3\sin(6x)-\sin(18x)dx$$ so $$\int\sin^3(3x)\cos^3(3x)dx=\frac{\cos(18x)}{576}-\frac{\cos(6x)}{192}+C$$ Oh, by parts...