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MarcLeclair
 one year ago
Integral question ( verifying my answer)
MarcLeclair
 one year ago
Integral question ( verifying my answer)

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MarcLeclair
 one year ago
Best ResponseYou've already chosen the best response.0\[\int\limits_{}^{} tsin(t)\cos(t)\] i did as follow: Integration by part u=t and du=dt / dv=sin(t)cos(t) v=sin^2(t)/2 which followed with \[(tsin ^{2}t)/2  \int\limits_{}^{}(\sin ^{2}t)/2dt \] then I get ( just for the integral) \[(1/4)\int\limits_{}^{}1\cos2t \rightarrow (1/4)(t+(\sin2t)/2)\] So my final answer looks like this: \[(tsin ^{2}t)/2  1/4(t+(\sin2t)/2 + c)\]

MarcLeclair
 one year ago
Best ResponseYou've already chosen the best response.0The answer key did it differently I just wanted to know if someone could help me confirm whether or not my answer looks right

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.1Hmm your process was a little strange. If you start by applying the `Double Angle Formula for Sine` it makes this problem a bit easier in my opinion. You end up with a solution of,\[\large \frac{1}{4}\cos(2t)+\frac{1}{8}\sin(2t)+C\] But looking at your solution, applying some identities, I can now see that it is equivalent. So it looks like it worked out for you! :)

MarcLeclair
 one year ago
Best ResponseYou've already chosen the best response.0Yeah it was a little strange to me, I just remembered a number where it used substitution within the integration by part. I honnestly kinda forgot how to use my substitution to prove a trig ( 3 years ago ) but alright thanks^^ It gets kinda hard to notice whether or not an answer is right when different methods can apply ahaha!

MarcLeclair
 one year ago
Best ResponseYou've already chosen the best response.0Yes the solution used double angle so I didn't come up as the same answer!
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