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anonymous
 3 years ago
Integral question ( verifying my answer)
anonymous
 3 years ago
Integral question ( verifying my answer)

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anonymous
 3 years 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: \[(tsin ^{2}t)/2  1/4(t+(\sin2t)/2 + c)\]

anonymous
 3 years 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
 3 years 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! :)

anonymous
 3 years 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!

anonymous
 3 years 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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