A community for students.
Here's the question you clicked on:
 0 viewing
UnkleRhaukus
 3 years ago
\[\frac23\int\limits_{0}^{3}\cos(x)\sin\left(\frac{\pi n x}{3/2}\right)\text dx\]
UnkleRhaukus
 3 years ago
\[\frac23\int\limits_{0}^{3}\cos(x)\sin\left(\frac{\pi n x}{3/2}\right)\text dx\]

This Question is Closed

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

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

UnkleRhaukus
 3 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
 3 years ago
Best ResponseYou've already chosen the best response.0funny my comment gt deleted !! anyways

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

anonymous
 3 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
 3 years ago
Best ResponseYou've already chosen the best response.2thanks for having a look,

anonymous
 3 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
 3 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
 3 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
 3 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
 3 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
 3 years ago
Best ResponseYou've already chosen the best response.2ah, i think thats it.

anonymous
 3 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
 3 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
 3 years ago
Best ResponseYou've already chosen the best response.0I need to sleep. good luck with it.

UnkleRhaukus
 3 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
 3 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
 3 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
 3 years ago
Best ResponseYou've already chosen the best response.2if you wanted some context,
Ask your own question
Sign UpFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.