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

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

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

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

stottrupbailey Group TitleBest ResponseYou've already chosen the best response.0
is that right?
 one year ago

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

LolWolf Group TitleBest ResponseYou've already chosen the best response.1
Yes, that is.
 one year ago

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

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

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

stottrupbailey Group TitleBest ResponseYou've already chosen the best response.0
dv=x dx
 one year ago

LolWolf Group TitleBest ResponseYou've already chosen the best response.1
Try 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.
 one year ago

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

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

LolWolf Group TitleBest ResponseYou've already chosen the best response.1
Quick 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.
 one year ago

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

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

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

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

stottrupbailey Group TitleBest ResponseYou've already chosen the best response.0
is that right?
 one year ago

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

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

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

LolWolf Group TitleBest ResponseYou've already chosen the best response.1
You'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.
 one year ago

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

LolWolf Group TitleBest ResponseYou've already chosen the best response.1
Sure thing, have a good one.
 one year ago
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