integrate 4-sin^3t

Question

integrate 4-sin^3t

jennychan12 · Answer

$$4-\sin^3t$$

jennychan12 · Answer

what i don't get is the sin^3t part

kainui · Answer

Are you supposed to actually integrate these, or approximate the integrals with riemann sums?

sin^3(t)=(sin(t))^3

jennychan12 · Answer

just integrate. i can do the rest.

jennychan12 · Answer

yeah i got that.

kainui · Answer

Ah, well I think you can solve this by separating it out into:

sint*sin^2tdt

and then from there use the identity sin^2t+cos^2=1

jennychan12 · Answer

ohh i see so sint(1-cos^2t)
so then sint-sintcos^2t ?
so then -cost...?

kainui · Answer

I think you're on the right path, yep.

jennychan12 · Answer

how about the -sintcos^2t how would you take the antiderivative of that?

kainui · Answer

Ahh, what would you normally do to solve an integral if you see a function and its derivative next to it?

jennychan12 · Answer

it's just the whatever the integral derivative thingy is...

kainui · Answer

Try u-substitution out. =D

kainui · Answer

I don't mean to be difficult, but integration is an art that takes practice. Giving you the answers too easily would be cheating you of thinking through it yourself.

kainui · Answer

Were you able to get it?

jennychan12 · Answer

oh yeah. i got 4t+cost-1/3cos^3t +C

kainui · Answer

I don't think that's quite right. I believe there's a + sign in front of 1/3cos^3t

jennychan12 · Answer

so i got the signs backwards?

kainui · Answer

Just on that one. I think you forgot that the integral looked like this:

$$-\int\limits_{?}^{?}sint-sintcos^2tdt$$

the negative goes away in front of the sint because the integral is -cost but it's also distributed out to the other part of the integral where the negatives cancel out.

jennychan12 · Answer

this is what my friend showed me

kainui · Answer

Actually you're right, I'm wrong. I think I put the derivative of cosine as sine, not negative sine.

jennychan12 · Answer

but my answer's wrong. i checked http://www.wolframalpha.com/input/?i=integrate+4-sin%5E3t

anonymous · Answer

You can use this if you want :

$$\sin(3x) = 3\sin(x) - 4 \sin^3(x)$$

anonymous · Answer

From here first find sin^3(x)..

kainui · Answer

I checked: http://www.wolframalpha.com/input/?i=4t%2B3%2F4cost-1%2F12cos%283t%29%3D4t%2Bcost-1%2F3cos%5E3t

jennychan12 · Answer

@Kainui why are we getting different answer on wolfram alpha?

jennychan12 · Answer

jk i see

anonymous · Answer

$$\sin^3(x) = \frac{3 \sin(x) - \sin(3x)}{4}$$

kainui · Answer

They used a reduction formula, which is something you won't use in calculus1, but you will in calculus 2. There is more than one way to get to the right answer, but the only reason you see that is because they use something that's more general and can be used for things like sin^8(t) which is unnecessary in this case.

anonymous · Answer

Do you think someone will give you sin^8(t) ??

jennychan12 · Answer

oh jeez.
please no.

kainui · Answer

I've seen very similar things in calculus 2. That's what reduction formulas are for. I could even derive them for you if you like. http://archives.math.utk.edu/visual.calculus/4/recursion.2/

jennychan12 · Answer

oyyyy...
i won't have to take calc bc (2)