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C4 Integration Question

This is the question

Integral[(sin 2t)(cos t) dt]

I tried doing this numerous times but I can never get rid of the multiplication between two functions. Its always goes to (sin 2t)(cos t) or (sin t)(cos 2t -1) which multiplies to get (sin t.cos 2t - sin t)
See, I can't get rid of the multiplication so how can I integrate it?

Any help appreciated!
Reply 1
Avatar for DFranklin
DFranklin
18
http://en.wikipedia.org/wiki/List_of_trigonometric_identities#Product-to-sum_and_sum-to-product_identities
Reply 2
Avatar for _bonnie
_bonnie
5
Not sure if this is right... but
Convert Sin2t = 2SintCost
and then see what you can do that way.
Without a Shadow
EDIT: Im wrong


'Fraid so :frown:

OP, try using integration by substitution.

Convert it using the double angle formula as the others have said, then use substitution.

u = cos x so du/dx = -sin x.

See it? :smile:
There are two approaches here:

1) The laborious way. If you integrate by parts twice, then the integrand you are left with will be in the same form as the original one, so you should be able to do it with a bit of algebraic manipulation.

2) The elegant and quick way (and possibly the way they are getting at in the question). Rewrite the sin 2t using your trusty double-angle formula. Then see if you can notice that the integrand happens to be a perfect derivative of something due to the chain rule.

Hope that helps, I've put the answer I got as a spoiler for you:

Spoiler

Reply 5
Avatar for aster100
aster100
sphixter
This is the question

Integral[(sin 2t)(cos t) dt]

I tried doing this numerous times but I can never get rid of the multiplication between two functions. Its always goes to (sin 2t)(cos t) or (sin t)(cos 2t -1) which multiplies to get (sin t.cos 2t - sin t)
See, I can't get rid of the multiplication so how can I integrate it?

Any help appreciated!


sin(2t+t) = sin2tcost + cos2tsint
sin(2t-t) = sin2tcost - cos2tsint

Add these together:

sin3t + sint = 2sin2tcost

sin2tcost = 1/2(sin3t + sint)

SAFE.

EDIT, oh sorry then integrating it is piss easy. -1/6cos3t - 1/2cost
Reply 6
Avatar for _bonnie
_bonnie
5
rob_the_sax_dude
Hope that helps, I've put the answer I got as a spoiler for you:


Hi, i got something different, so i wonder if you would be able to tell me where i went wrong?


∫(sin2t)(cost) dt[br][br]=∫(2sintcost)(cost) dt[br][br]=∫(2sintcos2t) dt[br][br]=∫(2sint(1−sin2t) dt[br][br]=∫(2sint−2sin3t) dt[br][br]=−2cost+1/2cost4t+c[br]\int (sin2t)(cost) \, dt [br][br]= \int (2sintcost)(cost) \, dt[br][br]= \int (2sintcos^2t) \, dt[br][br]= \int (2sint(1-sin^2t) \, dt[br][br]= \int (2sint - 2sin^3t) \, dt[br][br]= -2cost + 1/2cost^4t + c[br]
Reply 7
Avatar for aster100
aster100
[QUOTE="_bonnie"]Hi, i got something different, so i wonder if you would be able to tell me where i went wrong?


∫(sin2t)(cost) dt[br][br]=∫(2sintcost)(cost) dt[br][br]=∫(2sintcos2t) dt[br][br]=∫(2sint(1−sin2t) dt[br][br]=∫(2sint−2sin3t) dt[br][br]=−2cost+1/2cost4t+c[br][br][br]e.g.[latex]sin⁡2θ+cos⁡2θ=1[/latex]\int (sin2t)(cost) \, dt [br][br]= \int (2sintcost)(cost) \, dt[br][br]= \int (2sintcos^2t) \, dt[br][br]= \int (2sint(1-sin^2t) \, dt[br][br]= \int (2sint - 2sin^3t) \, dt[br][br]= -2cost + 1/2cost^4t + c[br][br][br]e.g. [latex]\sin^2 \theta + \cos^2 \theta=1[/latex]

Most probably the same thing.

If you see my solution above, it is also different.
Reply 8
Avatar for kabbers
kabbers
4
_bonnie
Hi, i got something different, so i wonder if you would be able to tell me where i went wrong?


∫(sin2t)(cost) dt[br][br]=∫(2sintcost)(cost) dt[br][br]=∫(2sintcos2t) dt[br][br]=∫(2sint(1−sin2t) dt[br][br]=∫(2sint−2sin3t) dt[br][br]=−2cost+1/2cost4t+c[br]\int (sin2t)(cost) \, dt [br][br]= \int (2sintcost)(cost) \, dt[br][br]= \int (2sintcos^2t) \, dt[br][br]= \int (2sint(1-sin^2t) \, dt[br][br]= \int (2sint - 2sin^3t) \, dt[br][br]= -2cost + 1/2cost^4t + c[br]


=∫(2sint−2sin3t) dt[br][br]=−2cost+1/2cost4t+c= \int (2sint - 2sin^3t) \, dt[br][br]= -2cost + 1/2cost^4t + c

that is what i disagree with, you cant integrate powers like that. instead leave the fiorst line as sin x cos^2 x, thensubstitute u = cos x.
Reply 9
Avatar for _bonnie
_bonnie
5
:redface: of course you can't, opps!
thanks for your help :smile:
Reply 10
Avatar for sphixter
sphixter
OP
aster100
sin(2t+t) = sin2tcost + cos2tsint
sin(2t-t) = sin2tcost - cos2tsint

Add these together:

sin3t + sint = 2sin2tcost

sin2tcost = 1/2(sin3t + sint)

SAFE.

EDIT, oh sorry then integrating it is piss easy. -1/6cos3t - 1/2cost


Thanks! Can't believe I didn't think of that.

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