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

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

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

is that right?

Yes, that is.

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 ?

dv=x dx

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

oh, gotcha. okay just a sec

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

is that right so far?

is that right?

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

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

Sure thing, have a good one.