Question

integrate xcos^2(8x)dx

Answer 1

\[\int\limits_{}^{}xcos^2(8x)dx\]

Answer 2

I think I might need a half angle formula? I'm still confused though...

Answer 3

Yep, power-reducing formula, and then integration by parts. Use the fact that:
\[
\cos^2(x)=\frac{1+\cos(2x)}{2}
\]

Answer 4

okay, so I've got it down to \[\int\limits_{}^{}x \frac{ 1 }{ 2 }(1+\cos(16x))dx\]

Answer 5

is that right?

Answer 6

To make life easier, think of it in terms of powers of e:
\[
e^{i\theta}=\cos(\theta)+i\sin(\theta)
\]If you wish.

Answer 7

Yes, that is.

Answer 8

can you give me a hint what to do next please?

Answer 9

Sure: try to integrate it using integration by parts. Do you know how to do this...?

Answer 10

Yes I do. Should I use u=1+cos(16x) and dv=x ?

Answer 11

dv=x dx

Answer 12

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.

Answer 13

Try doing the following case:
\[
dv=(1+\cos(16))dx,\;v=\cdots\\
u=x,\;du=dx
\]

Answer 14

oh, gotcha. okay just a sec

Answer 15

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.

Answer 16

\[x(x+\frac{ \sin(16x) }{ 16 })-\int\limits_{}^{}x+\frac{ \sin(16x) }{ 16 } dx\]

Answer 17

is that right so far?

Answer 18

Yessir/ma'am.

Although don't forget the constant \(\frac{1}{2}\) that was at the beginning of the problem.

Answer 19

yep I left that hanging around somewhere. Okay, so without the 1/2 from before I got \[x^2+xsin(16x)/16 -\int\limits_{}^{}x+\frac{ \sin(16x) }{ 16 }dx\]

Answer 20

is that right?

Answer 21

which would then be \[x^2+\frac{ xsin(16x) }{ 16 }-\frac{ x^2 }{ 2 }-\frac{ \cos(16x) }{ 32 }\]

Answer 22

oops that last one should be plus, not minus

Answer 23

ahhh nope that's not right just a sec

Answer 24

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.

Answer 25

Okay, I got the right answer. Thanks so much for the help!! :)

Answer 26

Sure thing, have a good one.