Hey all-- I had an Integration of Parts quiz today and got stuck on a few problems-- was wondering if you could explain the steps involved in solving these integrals: int( (sin(3x))^3 * (cos(3x))^3 dx) and int( (tan(4x))^4) dx) thanks!!
[tex]\int \sin^3{3x} \cos^3{3x}\,dx[/tex] 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, [tex]\frac{1}{3}\int \sin^3{u} \cos^3{u}\,du[/tex] 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! [tex]\sin^2{x} + \cos^2{x} = 1[/tex] Use this to turn the integral into [tex]\frac{1}{3}\int \sin^3{u} (1 - \sin^2{u}) \cos{u}\,du[/tex] which is easily separated and solved by substitution. Have another shot at the second one, keeping in mind that [tex]\frac{d}{dx}\tan{x} = \sec^2{x}[/tex] cookiemonster
Where did you get [tex]\cos^3=1-\sin^2[/tex] I would probably use the half angle formulas: [tex]\sin(2x)=2\sin(x)\cos(x)[/tex] so [tex]\sin(x)\cos(x)=\frac{\sin(2x)}{2}[/tex] so [tex]\sin(3x)\cos(3x)=\frac{\sin(6x)}{2}[/tex] Now you've got: [tex]\int\sin^3(3x)\cos^3(3x)dx[/tex] [tex]\int(\sin(3x)\cos(3x))^3dx[/tex] [tex]\frac{1}{8}\int\sin^3(6x)dx[/tex] Now [tex]\sin(3x)=3\sin(x)-4\sin^3(x)[/tex] so [tex]\sin^3(6x)=\frac{3\sin(6x)-\sin(18x)}{4}[/tex] so [tex]\frac{1}{32}\int3\sin(6x)-\sin(18x)dx[/tex] so [tex]\int\sin^3(3x)\cos^3(3x)dx=\frac{\cos(18x)}{576}-\frac{\cos(6x)}{192}+C[/tex] Oh, by parts...
I didn't. I only took two of the cosines and I left the third for the u substitution. u = sinx du = cosxdx It's used in the du. Edit: By parts? cookiemonster