Question

Could someone please check my answer to the following integral (click to see).

Answer 1

\[\int\limits_{}^{}\sin ^{3}(a \theta)d \theta = -\frac{ 2\cos(a \theta) }{ 3a }+C\]

Answer 2

i don't think thats correct ...how you got that anyways ?

Answer 3

for a hint, you need to use the formula for sin 3theta haere.

Answer 4

I broke off a sin^2(atheta) and used u=cos(atheta)

Answer 5

That is the right way to do it, so you got\[\int\limits \sin^3(a \theta)d \theta=\int\limits \sin^2(a \theta)\sin(a \theta) d \theta=\int\limits (1-\cos^2(a \theta))\sin(a \theta) d \theta\]Now, if you use \(u=\cos(a \theta)\), then \(du=-asin(a \theta)\), so you have:\[-\frac{1}{a} \int\limits(1-u^2)du\]This equals\[- \frac{1}{a}(u- \frac{1}{3}u^3)+C\]

Answer 6

yea that's what i did

Answer 7

oh ok i found my mistake, thanks...

Answer 8

ahh i still got it wrong...

Answer 9

Now we have to substitute back \(u=cos(a \theta)\):\[- \frac{1}{a}(\cos(a \theta)- \frac{1}{3}\cos^3(a \theta))+C\]After a few steps, this is what I get:\[-\frac{\cos(a \theta)(2+\sin^2(a \theta))}{3a}+C\]So, I do not get you answer...

Answer 10

my answer now is...
\[\frac{ \cos^3(a \theta) }{ 3a }-\frac{ \cos(a \theta) }{ a }+C\]

Answer 11

I think that is the same as mine...

Answer 12

you didn't multiply (-1/3u^3) by (-1/a)

Answer 13

Thought I did, see 4 steps up ^^

Answer 14

the answer in the book is
\[\frac{ \cos^3(a \theta) }{ 3a }-\cos (a \theta)+C\]

Answer 15

They forgot to divide cos(a theta) by a.