- #1
Saladsamurai
- 3,020
- 7
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
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
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