Open study

is now brainly

With Brainly you can:

  • Get homework help from millions of students and moderators
  • Learn how to solve problems with step-by-step explanations
  • Share your knowledge and earn points by helping other students
  • Learn anywhere, anytime with the Brainly app!

A community for students.

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

Mathematics
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this and thousands of other questions

@UnkleRhaukus whats in ur mind ?
i tried that , but i keep getting the wrong answer
\[\begin{align*} % b_n b_n&=\frac23\int\limits_{0}^{3}\cos(x)\sin\left(\frac{\pi n x}{3/2}\right)\text dx\\ &=\frac13\int\limits_{0}^{3}\sin\left((\tfrac23\pi n+1)x\right)-\sin\left((\tfrac23\pi n-1)x\right)\text dx\\\ &=-\frac13\left.\left(\frac{\cos\left((\tfrac23\pi n+1)x\right)}{\tfrac23\pi n+1}-\frac{\cos\left((\tfrac23\pi n-1)x\right)}{\tfrac23\pi n-1}\right)\right|_0^3\\ &=\frac{1-\cos\left(2\pi n+3\right)}{2\pi n+3}-\frac{1-\cos\left(2\pi n-3\right)}{2\pi n-3}\\ &=\frac{1-\cos\left(3\right)}{2\pi n+3}-\frac{1-\cos\left(3\right)}{2\pi n-3}\\ &=\left(1-\cos\left(3\right)\right)\left(\frac{1}{2\pi n+3}-\frac{1}{2\pi n-3}\right)\\ &=\left(1-\cos\left(3\right)\right)\left(\frac{2\pi n-3}{(2\pi n+3)(2\pi n-3)}-\frac{2\pi n+3}{(2\pi n-3)(2\pi n+3)}\right)\\ &=\left(1-\cos\left(3\right)\right)\left(\frac{-6}{4\pi^2n^2-9}\right)\\ \end{align*}\]

Not the answer you are looking for?

Search for more explanations.

Ask your own question

Other answers:

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.
thank you
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,
1 Attachment

Not the answer you are looking for?

Search for more explanations.

Ask your own question