Here's the question you clicked on:
stottrupbailey
integrate xcos^2(8x)dx
\[\int\limits_{}^{}xcos^2(8x)dx\]
I think I might need a half angle formula? I'm still confused though...
Yep, power-reducing formula, and then integration by parts. Use the fact that: \[ \cos^2(x)=\frac{1+\cos(2x)}{2} \]
okay, so I've got it down to \[\int\limits_{}^{}x \frac{ 1 }{ 2 }(1+\cos(16x))dx\]
To make life easier, think of it in terms of powers of e: \[ e^{i\theta}=\cos(\theta)+i\sin(\theta) \]If you wish.
can you give me a hint what to do next please?
Sure: try to integrate it using integration by parts. Do you know how to do this...?
Yes I do. Should I use u=1+cos(16x) and dv=x ?
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.
Try doing the following case: \[ dv=(1+\cos(16))dx,\;v=\cdots\\ u=x,\;du=dx \]
oh, gotcha. okay just a sec
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.
\[x(x+\frac{ \sin(16x) }{ 16 })-\int\limits_{}^{}x+\frac{ \sin(16x) }{ 16 } dx\]
is that right so far?
Yessir/ma'am. Although don't forget the constant \(\frac{1}{2}\) that was at the beginning of the problem.
yep I left that hanging around somewhere. Okay, so without the 1/2 from before I got \[x^2+xsin(16x)/16 -\int\limits_{}^{}x+\frac{ \sin(16x) }{ 16 }dx\]
which would then be \[x^2+\frac{ xsin(16x) }{ 16 }-\frac{ x^2 }{ 2 }-\frac{ \cos(16x) }{ 32 }\]
oops that last one should be plus, not minus
ahhh nope that's not right just a sec
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.
Okay, I got the right answer. Thanks so much for the help!! :)
Sure thing, have a good one.