A community for students. Sign up today!
Here's the question you clicked on:
 0 viewing

This Question is Closed

stottrupbailey
 one year ago
Best ResponseYou've already chosen the best response.0\[\int\limits_{}^{}xcos^2(8x)dx\]

stottrupbailey
 one year ago
Best ResponseYou've already chosen the best response.0I think I might need a half angle formula? I'm still confused though...

LolWolf
 one year ago
Best ResponseYou've already chosen the best response.1Yep, powerreducing formula, and then integration by parts. Use the fact that: \[ \cos^2(x)=\frac{1+\cos(2x)}{2} \]

stottrupbailey
 one year ago
Best ResponseYou've already chosen the best response.0okay, so I've got it down to \[\int\limits_{}^{}x \frac{ 1 }{ 2 }(1+\cos(16x))dx\]

LolWolf
 one year ago
Best ResponseYou've already chosen the best response.1To make life easier, think of it in terms of powers of e: \[ e^{i\theta}=\cos(\theta)+i\sin(\theta) \]If you wish.

stottrupbailey
 one year ago
Best ResponseYou've already chosen the best response.0can you give me a hint what to do next please?

LolWolf
 one year ago
Best ResponseYou've already chosen the best response.1Sure: try to integrate it using integration by parts. Do you know how to do this...?

stottrupbailey
 one year ago
Best ResponseYou've already chosen the best response.0Yes I do. Should I use u=1+cos(16x) and dv=x ?

LolWolf
 one year ago
Best ResponseYou've already chosen the best response.1Try the other way around. Since we wish to reduce it to an integrable series. E.g. we can integrate \(\sin(l)\), but not \(\frac{x^2}{2}\sin(l)\), without integration by parts.

LolWolf
 one year ago
Best ResponseYou've already chosen the best response.1Try doing the following case: \[ dv=(1+\cos(16))dx,\;v=\cdots\\ u=x,\;du=dx \]

stottrupbailey
 one year ago
Best ResponseYou've already chosen the best response.0oh, gotcha. okay just a sec

LolWolf
 one year ago
Best ResponseYou've already chosen the best response.1Quick tip, remember, intuitively, what integration by parts does: For some functions \(f, g\), integration by parts turns the problem: \[ \int f'\cdot g\;dx \]Into the problem: \[ \int f\cdot g'\;dx \]Think about this a little bit and it will become much easier to understand its use.

stottrupbailey
 one year ago
Best ResponseYou've already chosen the best response.0\[x(x+\frac{ \sin(16x) }{ 16 })\int\limits_{}^{}x+\frac{ \sin(16x) }{ 16 } dx\]

stottrupbailey
 one year ago
Best ResponseYou've already chosen the best response.0is that right so far?

LolWolf
 one year ago
Best ResponseYou've already chosen the best response.1Yessir/ma'am. Although don't forget the constant \(\frac{1}{2}\) that was at the beginning of the problem.

stottrupbailey
 one year ago
Best ResponseYou've already chosen the best response.0yep I left that hanging around somewhere. Okay, so without the 1/2 from before I got \[x^2+xsin(16x)/16 \int\limits_{}^{}x+\frac{ \sin(16x) }{ 16 }dx\]

stottrupbailey
 one year ago
Best ResponseYou've already chosen the best response.0which would then be \[x^2+\frac{ xsin(16x) }{ 16 }\frac{ x^2 }{ 2 }\frac{ \cos(16x) }{ 32 }\]

stottrupbailey
 one year ago
Best ResponseYou've already chosen the best response.0oops that last one should be plus, not minus

stottrupbailey
 one year ago
Best ResponseYou've already chosen the best response.0ahhh nope that's not right just a sec

LolWolf
 one year ago
Best ResponseYou've already chosen the best response.1You're really close, remember that the denominator has to multiply by 16, not add, and that the integral of sin is cos, not +cos. But, you've got the jist. The devil is in the details.

stottrupbailey
 one year ago
Best ResponseYou've already chosen the best response.0Okay, I got the right answer. Thanks so much for the help!! :)

LolWolf
 one year ago
Best ResponseYou've already chosen the best response.1Sure thing, have a good one.
Ask your own question
Ask a QuestionFind 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.