## stottrupbailey 2 years ago integrate xcos^2(8x)dx

1. stottrupbailey

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

2. stottrupbailey

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

3. LolWolf

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

4. stottrupbailey

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

5. stottrupbailey

is that right?

6. LolWolf

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

7. LolWolf

Yes, that is.

8. stottrupbailey

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

9. LolWolf

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

10. stottrupbailey

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

11. stottrupbailey

dv=x dx

12. LolWolf

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.

13. LolWolf

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

14. stottrupbailey

oh, gotcha. okay just a sec

15. LolWolf

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.

16. stottrupbailey

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

17. stottrupbailey

is that right so far?

18. LolWolf

Yessir/ma'am. Although don't forget the constant $$\frac{1}{2}$$ that was at the beginning of the problem.

19. stottrupbailey

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

20. stottrupbailey

is that right?

21. stottrupbailey

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

22. stottrupbailey

oops that last one should be plus, not minus

23. stottrupbailey

ahhh nope that's not right just a sec

24. LolWolf

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.

25. stottrupbailey

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

26. LolWolf

Sure thing, have a good one.