Here's the question you clicked on:
UnkleRhaukus
\[\frac23\int\limits_{0}^{3}\cos(x)\sin\left(\frac{\pi n x}{3/2}\right)\text dx\]
@UnkleRhaukus whats in ur mind ?
i tried that , but i keep getting the wrong answer
\[\begin{align*} % b_n b_n&=\frac23\int\limits_{0}^{3}\cos(x)\sin\left(\frac{\pi n x}{3/2}\right)\text dx\\ &=\frac13\int\limits_{0}^{3}\sin\left((\tfrac23\pi n+1)x\right)-\sin\left((\tfrac23\pi n-1)x\right)\text dx\\\ &=-\frac13\left.\left(\frac{\cos\left((\tfrac23\pi n+1)x\right)}{\tfrac23\pi n+1}-\frac{\cos\left((\tfrac23\pi n-1)x\right)}{\tfrac23\pi n-1}\right)\right|_0^3\\ &=\frac{1-\cos\left(2\pi n+3\right)}{2\pi n+3}-\frac{1-\cos\left(2\pi n-3\right)}{2\pi n-3}\\ &=\frac{1-\cos\left(3\right)}{2\pi n+3}-\frac{1-\cos\left(3\right)}{2\pi n-3}\\ &=\left(1-\cos\left(3\right)\right)\left(\frac{1}{2\pi n+3}-\frac{1}{2\pi n-3}\right)\\ &=\left(1-\cos\left(3\right)\right)\left(\frac{2\pi n-3}{(2\pi n+3)(2\pi n-3)}-\frac{2\pi n+3}{(2\pi n-3)(2\pi n+3)}\right)\\ &=\left(1-\cos\left(3\right)\right)\left(\frac{-6}{4\pi^2n^2-9}\right)\\ \end{align*}\]
funny my comment gt deleted !! anyways
am i using the wrong trig substitution?
sorry I'm not very good at the trigonometric identities... I'm a bit lost in your derivation, due to my own limits.
thanks for having a look,
http://www.sosmath.com/trig/Trig5/trig5/trig5.html Are you using the second to last 'product to sum' formula here?
i used the last product to sum formula, would the result be different if i used the second to last formula/?
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.
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...
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?
ah, i think thats it.
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...)
I just typed in the indefinite integral form, to see how messy the anti-derivative is. Then you would need to evaluate at x=0 and x=3 in the usual way.
I need to sleep. good luck with it.
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 \[\begin{align*} % b_n b_n&=\frac23\int\limits_{0}^{3}\cos(x)\sin\left(\frac{\pi n x}{3/2}\right)\text dx\\ &=\frac13\int\limits_{0}^{3}\sin\left((1+\tfrac23\pi n)x\right)-\sin\left((1-\tfrac23\pi n)x\right)\text dx\\\ &=-\frac13\left.\left(\frac{\cos\left((1+\tfrac23\pi n)x\right)}{1+\tfrac23\pi n}-\frac{\cos\left((1-\tfrac23\pi n)x\right)}{1-\tfrac23\pi n}\right)\right|_0^3\\ &=\frac{1-\cos\left(3+2\pi n\right)}{2\pi n+3}-\frac{1-\cos\left(3-2\pi n\right)}{2\pi n-3}\\ &=\frac{1-\cos\left(3\right)}{3+2\pi n}-\frac{1-\cos\left(3\right)}{3-2\pi n}\\ &=\left(1-\cos\left(3\right)\right)\left(\frac{1}{3+2\pi n}-\frac{1}{3-2\pi n}\right)\\ &=\left(1-\cos\left(3\right)\right)\left(\frac{3-2\pi n}{(3+2\pi n)(3-2\pi n)}-\frac{3+2\pi n}{(3-2\pi n)(3+2\pi n)}\right)\\ &=\left(1-\cos\left(3\right)\right)\left(\frac{-4\pi n}{9-4\pi^2n^2}\right)\\ &=\left(1-\cos\left(3\right)\right)\left(\frac{4\pi n}{4\pi^2n^2-9}\right)\color{red} \checkmark\\ \end{align*}\]
cool. That looks reasonable! Well done. I didn't help that much. I'm terrible at these trigonometric identities...
It's great that you caught that little detail. It's always the little details... :P
if you wanted some context,