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\[\frac23\int\limits_{0}^{3}\cos(x)\sin\left(\frac{\pi n x}{3/2}\right)\text dx\]
 one year ago
 one year ago
\[\frac23\int\limits_{0}^{3}\cos(x)\sin\left(\frac{\pi n x}{3/2}\right)\text dx\]
 one year ago
 one year ago

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AravindGBest ResponseYou've already chosen the best response.0
@UnkleRhaukus whats in ur mind ?
 one year ago

UnkleRhaukusBest ResponseYou've already chosen the best response.2
i tried that , but i keep getting the wrong answer
 one year ago

UnkleRhaukusBest ResponseYou've already chosen the best response.2
\[\begin{align*} % b_n b_n&=\frac23\int\limits_{0}^{3}\cos(x)\sin\left(\frac{\pi n x}{3/2}\right)\text dx\\ &=\frac13\int\limits_{0}^{3}\sin\left((\tfrac23\pi n+1)x\right)\sin\left((\tfrac23\pi n1)x\right)\text dx\\\ &=\frac13\left.\left(\frac{\cos\left((\tfrac23\pi n+1)x\right)}{\tfrac23\pi n+1}\frac{\cos\left((\tfrac23\pi n1)x\right)}{\tfrac23\pi n1}\right)\right_0^3\\ &=\frac{1\cos\left(2\pi n+3\right)}{2\pi n+3}\frac{1\cos\left(2\pi n3\right)}{2\pi n3}\\ &=\frac{1\cos\left(3\right)}{2\pi n+3}\frac{1\cos\left(3\right)}{2\pi n3}\\ &=\left(1\cos\left(3\right)\right)\left(\frac{1}{2\pi n+3}\frac{1}{2\pi n3}\right)\\ &=\left(1\cos\left(3\right)\right)\left(\frac{2\pi n3}{(2\pi n+3)(2\pi n3)}\frac{2\pi n+3}{(2\pi n3)(2\pi n+3)}\right)\\ &=\left(1\cos\left(3\right)\right)\left(\frac{6}{4\pi^2n^29}\right)\\ \end{align*}\]
 one year ago

AravindGBest ResponseYou've already chosen the best response.0
funny my comment gt deleted !! anyways
 one year ago

UnkleRhaukusBest ResponseYou've already chosen the best response.2
am i using the wrong trig substitution?
 one year ago

scarydoorBest ResponseYou've already chosen the best response.2
sorry I'm not very good at the trigonometric identities... I'm a bit lost in your derivation, due to my own limits.
 one year ago

UnkleRhaukusBest ResponseYou've already chosen the best response.2
thanks for having a look,
 one year ago

scarydoorBest ResponseYou've already chosen the best response.2
http://www.sosmath.com/trig/Trig5/trig5/trig5.html Are you using the second to last 'product to sum' formula here?
 one year ago

UnkleRhaukusBest ResponseYou've already chosen the best response.2
i used the last product to sum formula, would the result be different if i used the second to last formula/?
 one year ago

scarydoorBest ResponseYou've already chosen the best response.2
not sure. I might have just made a mistake in thikning it was the second to last one. The second to last one might not even apply properly.
 one year ago

AccessDeniedBest ResponseYou've already chosen the best response.0
I think they'd both be the 'same,' except one was modified slightly with a negative factored out of sine to get it into a nicer form. The multiplication 'should' be commutative there...
 one year ago

scarydoorBest ResponseYou've already chosen the best response.2
On the second line, the second term in the integral you've got: \[sin[(\frac{2}{3} \pi n  1)x] \] whereas I've got it as: \[sin[(1\frac{2}{3}\pi n) x] \] Maybe you did something else to get your expression that I haven't seen. Or else maybe that's an error?
 one year ago

UnkleRhaukusBest ResponseYou've already chosen the best response.2
ah, i think thats it.
 one year ago

scarydoorBest ResponseYou've already chosen the best response.2
Wolfram Alpha thinks the solution is not nice looking: http://www.wolframalpha.com/input/?i=integral+of+1%2F3+*+cos+x+sin%28%282pi+n+x+%29+%2F+3%29 (can be useful to confirm that it's supposed to look pretty bad...)
 one year ago

scarydoorBest ResponseYou've already chosen the best response.2
I just typed in the indefinite integral form, to see how messy the antiderivative is. Then you would need to evaluate at x=0 and x=3 in the usual way.
 one year ago

scarydoorBest ResponseYou've already chosen the best response.2
I need to sleep. good luck with it.
 one year ago

UnkleRhaukusBest ResponseYou've already chosen the best response.2
wolfram is having trouble because it dosent realize n in an integer but ive got what i wanted now i have the negative sign fixed thanks again @scarydoor \[\begin{align*} % b_n b_n&=\frac23\int\limits_{0}^{3}\cos(x)\sin\left(\frac{\pi n x}{3/2}\right)\text dx\\ &=\frac13\int\limits_{0}^{3}\sin\left((1+\tfrac23\pi n)x\right)\sin\left((1\tfrac23\pi n)x\right)\text dx\\\ &=\frac13\left.\left(\frac{\cos\left((1+\tfrac23\pi n)x\right)}{1+\tfrac23\pi n}\frac{\cos\left((1\tfrac23\pi n)x\right)}{1\tfrac23\pi n}\right)\right_0^3\\ &=\frac{1\cos\left(3+2\pi n\right)}{2\pi n+3}\frac{1\cos\left(32\pi n\right)}{2\pi n3}\\ &=\frac{1\cos\left(3\right)}{3+2\pi n}\frac{1\cos\left(3\right)}{32\pi n}\\ &=\left(1\cos\left(3\right)\right)\left(\frac{1}{3+2\pi n}\frac{1}{32\pi n}\right)\\ &=\left(1\cos\left(3\right)\right)\left(\frac{32\pi n}{(3+2\pi n)(32\pi n)}\frac{3+2\pi n}{(32\pi n)(3+2\pi n)}\right)\\ &=\left(1\cos\left(3\right)\right)\left(\frac{4\pi n}{94\pi^2n^2}\right)\\ &=\left(1\cos\left(3\right)\right)\left(\frac{4\pi n}{4\pi^2n^29}\right)\color{red} \checkmark\\ \end{align*}\]
 one year ago

scarydoorBest ResponseYou've already chosen the best response.2
cool. That looks reasonable! Well done. I didn't help that much. I'm terrible at these trigonometric identities...
 one year ago

AccessDeniedBest ResponseYou've already chosen the best response.0
It's great that you caught that little detail. It's always the little details... :P
 one year ago

UnkleRhaukusBest ResponseYou've already chosen the best response.2
if you wanted some context,
 one year ago
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