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anonymous
 3 years ago
integrate xcos^2(8x)dx
anonymous
 3 years ago
integrate xcos^2(8x)dx

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

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Yep, powerreducing formula, and then integration by parts. Use the fact that: \[ \cos^2(x)=\frac{1+\cos(2x)}{2} \]

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0To make life easier, think of it in terms of powers of e: \[ e^{i\theta}=\cos(\theta)+i\sin(\theta) \]If you wish.

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Sure: try to integrate it using integration by parts. Do you know how to do this...?

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Try 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.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Try doing the following case: \[ dv=(1+\cos(16))dx,\;v=\cdots\\ u=x,\;du=dx \]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0oh, gotcha. okay just a sec

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Quick 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.

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0is that right so far?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Yessir/ma'am. Although don't forget the constant \(\frac{1}{2}\) that was at the beginning of the problem.

anonymous
 3 years 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 \[x^2+xsin(16x)/16 \int\limits_{}^{}x+\frac{ \sin(16x) }{ 16 }dx\]

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

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0ahhh nope that's not right just a sec

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0You'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.

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Sure thing, have a good one.
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