Okay so I am suposed to evaluate 2 of these: 1.) [tex]\int\sin^3(a\theta)d\theta[/tex] the solution manual looks like it used a trig ID to do this. Why?? Doesn't [tex]\int\sin^3u du=\frac{1}{3}\cos^3u-(\cos u)+C[/tex] ? So just use the u-sub [tex]u=a\theta \Rightarrow du/a=d\theta[/tex] So it should just be [tex]\frac{1}{3a}\cos^3a\theta-\frac{1}{a}\cosa\theta+C[/tex] ? But they got: [tex]-\frac{1}{a}\cos a\theta-\frac{1}{3a}\cos^3a\theta+C[/tex] Why the negative -1/3a ? What did I miss? Is it my formula? 2.) For [tex]\int\sin^4(3x)\cos^3xdx[/tex] they went from <---that to [tex]=\int\sin^4(3x)(1-sin^23x)\cos3xdx[/tex] How the hell does that make this problem ANY easier?! Is there a formula for ^^^that! Casey
You're using sin(u) like a variable, and integrating, which will certainly get you nowhere. One thing is that [tex] \int sin(u)^{3}[/tex] does not equal [tex]\frac{1}{3}\cos^3u[/tex]. You're going to have to use u-substitution, as you already have, and almost certainly some integration by parts.
I'll try that now. But why is my formula not producing the correct result if I am using u=a*theta ? Shouldn't it be 1/a *{Pug it in} ?
[tex]\int \sin^3(a\theta)d\theta = \int \sin(a\theta)d\theta - \int \sin(a\theta)\cos^2(a\theta)d\theta[/tex] The first term should be easy for you to integrate, for the second term, try a substitution of [tex]u = \cos(a\theta) \Rightarrow \frac{du}{d\theta} = -a\sin(a\theta) \Rightarrow \sin(a\theta)d\theta = -\frac{1}{a}du[/tex] After which the equation simplifies a little. Use a similar idea for the second one.
So I got it using this. But I am still trying to figure out where I went wrong with the formula. Casey
Seems to me like there is simply a typo in the book (the coefficient of the cos^3 should indeed be positive)
No, it's positive. (1-cos^(x)^2)*sin(x)*dx. u=cos(x), du=-sin(x)*dx. The u^3/3 term comes out with a plus sign.