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stottrupbailey

  • 3 years ago

integrate xcos^2(8x)dx

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  1. stottrupbailey
    • 3 years ago
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    \[\int\limits_{}^{}xcos^2(8x)dx\]

  2. stottrupbailey
    • 3 years ago
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    I think I might need a half angle formula? I'm still confused though...

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

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

  5. stottrupbailey
    • 3 years ago
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    is that right?

  6. LolWolf
    • 3 years ago
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    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
    • 3 years ago
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    Yes, that is.

  8. stottrupbailey
    • 3 years ago
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    can you give me a hint what to do next please?

  9. LolWolf
    • 3 years ago
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    Sure: try to integrate it using integration by parts. Do you know how to do this...?

  10. stottrupbailey
    • 3 years ago
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    Yes I do. Should I use u=1+cos(16x) and dv=x ?

  11. stottrupbailey
    • 3 years ago
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    dv=x dx

  12. LolWolf
    • 3 years ago
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    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
    • 3 years ago
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    Try doing the following case: \[ dv=(1+\cos(16))dx,\;v=\cdots\\ u=x,\;du=dx \]

  14. stottrupbailey
    • 3 years ago
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    oh, gotcha. okay just a sec

  15. LolWolf
    • 3 years ago
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    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
    • 3 years ago
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    \[x(x+\frac{ \sin(16x) }{ 16 })-\int\limits_{}^{}x+\frac{ \sin(16x) }{ 16 } dx\]

  17. stottrupbailey
    • 3 years ago
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    is that right so far?

  18. LolWolf
    • 3 years ago
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    Yessir/ma'am. Although don't forget the constant \(\frac{1}{2}\) that was at the beginning of the problem.

  19. stottrupbailey
    • 3 years ago
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    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
    • 3 years ago
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    is that right?

  21. stottrupbailey
    • 3 years ago
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    which would then be \[x^2+\frac{ xsin(16x) }{ 16 }-\frac{ x^2 }{ 2 }-\frac{ \cos(16x) }{ 32 }\]

  22. stottrupbailey
    • 3 years ago
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    oops that last one should be plus, not minus

  23. stottrupbailey
    • 3 years ago
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    ahhh nope that's not right just a sec

  24. LolWolf
    • 3 years ago
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    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
    • 3 years ago
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    Okay, I got the right answer. Thanks so much for the help!! :)

  26. LolWolf
    • 3 years ago
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    Sure thing, have a good one.

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