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

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AravindG
 4 years ago
Best ResponseYou've already chosen the best response.0@UnkleRhaukus whats in ur mind ?

UnkleRhaukus
 4 years ago
Best ResponseYou've already chosen the best response.2i tried that , but i keep getting the wrong answer

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

AravindG
 4 years ago
Best ResponseYou've already chosen the best response.0funny my comment gt deleted !! anyways

UnkleRhaukus
 4 years ago
Best ResponseYou've already chosen the best response.2am i using the wrong trig substitution?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0sorry I'm not very good at the trigonometric identities... I'm a bit lost in your derivation, due to my own limits.

UnkleRhaukus
 4 years ago
Best ResponseYou've already chosen the best response.2thanks for having a look,

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0http://www.sosmath.com/trig/Trig5/trig5/trig5.html Are you using the second to last 'product to sum' formula here?

UnkleRhaukus
 4 years ago
Best ResponseYou've already chosen the best response.2i used the last product to sum formula, would the result be different if i used the second to last formula/?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0not 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.

AccessDenied
 4 years ago
Best ResponseYou've already chosen the best response.0I 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...

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0On 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?

UnkleRhaukus
 4 years ago
Best ResponseYou've already chosen the best response.2ah, i think thats it.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Wolfram 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...)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I 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.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I need to sleep. good luck with it.

UnkleRhaukus
 4 years ago
Best ResponseYou've already chosen the best response.2wolfram 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 \[\begin{align*} % b_n b_n&=\frac23\int\limits_{0}^{3}\cos(x)\sin\left(\frac{\pi n x}{3/2}\right)\text dx\\ &=\frac13\int\limits_{0}^{3}\sin\left((1+\tfrac23\pi n)x\right)\sin\left((1\tfrac23\pi n)x\right)\text dx\\\ &=\frac13\left.\left(\frac{\cos\left((1+\tfrac23\pi n)x\right)}{1+\tfrac23\pi n}\frac{\cos\left((1\tfrac23\pi n)x\right)}{1\tfrac23\pi n}\right)\right_0^3\\ &=\frac{1\cos\left(3+2\pi n\right)}{2\pi n+3}\frac{1\cos\left(32\pi n\right)}{2\pi n3}\\ &=\frac{1\cos\left(3\right)}{3+2\pi n}\frac{1\cos\left(3\right)}{32\pi n}\\ &=\left(1\cos\left(3\right)\right)\left(\frac{1}{3+2\pi n}\frac{1}{32\pi n}\right)\\ &=\left(1\cos\left(3\right)\right)\left(\frac{32\pi n}{(3+2\pi n)(32\pi n)}\frac{3+2\pi n}{(32\pi n)(3+2\pi n)}\right)\\ &=\left(1\cos\left(3\right)\right)\left(\frac{4\pi n}{94\pi^2n^2}\right)\\ &=\left(1\cos\left(3\right)\right)\left(\frac{4\pi n}{4\pi^2n^29}\right)\color{red} \checkmark\\ \end{align*}\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0cool. That looks reasonable! Well done. I didn't help that much. I'm terrible at these trigonometric identities...

AccessDenied
 4 years ago
Best ResponseYou've already chosen the best response.0It's great that you caught that little detail. It's always the little details... :P

UnkleRhaukus
 4 years ago
Best ResponseYou've already chosen the best response.2if you wanted some context,
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