Okay so I am suposed to evaluate 2 of these:(adsbygoogle = window.adsbygoogle || []).push({});

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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# Integrating Powers of sin

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