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dfresenius Group Title

Integral with trig functions

  • one year ago
  • one year ago

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  1. dfresenius Group Title
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    |dw:1361089150658:dw|

    • one year ago
  2. dumbcow Group Title
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    i am intrigued

    • one year ago
  3. dfresenius Group Title
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    |dw:1361089181888:dw|

    • one year ago
  4. dfresenius Group Title
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    or would it be cos^xcos5x

    • one year ago
  5. dumbcow Group Title
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    no cant combine them...it would just be \[\cos(5x) - \cos^{2} x \cos(5x)\]

    • one year ago
  6. dfresenius Group Title
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    ok

    • one year ago
  7. dfresenius Group Title
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    so left integral is easy

    • one year ago
  8. dfresenius Group Title
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    hmm no i cant do that

    • one year ago
  9. dfresenius Group Title
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    that 5x is just blowing my mind

    • one year ago
  10. dumbcow Group Title
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    yeah im not sure of way to do it without expanding out the cos(5x) ..... which would be messy

    • one year ago
  11. hartnn Group Title
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    huh ? integral of cos 5x is just sin 5x/5 +c .

    • one year ago
  12. dfresenius Group Title
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    |dw:1361089595564:dw|

    • one year ago
  13. dfresenius Group Title
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    yA i know that integral, but its the cos^2 that is messin me up

    • one year ago
  14. hartnn Group Title
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    ok, you need to used \(\sin^2x = (1- \cos2x )/2\)

    • one year ago
  15. dfresenius Group Title
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    okkk

    • one year ago
  16. hartnn Group Title
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    then, for cos 2x cos 7x there is a standard formula for , cos A cos B =...

    • one year ago
  17. hartnn Group Title
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    *cos 2x cos 5x

    • one year ago
  18. dumbcow Group Title
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    ahh i always forget sum to product rules ... anyway, here is solution http://www.wolframalpha.com/input/?i=integrate+sin^2%28x%29+cos%285x%29+dx

    • one year ago
  19. dfresenius Group Title
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    i got it then :)

    • one year ago
  20. hartnn Group Title
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    good!

    • one year ago
  21. dfresenius Group Title
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    |dw:1361090139384:dw|

    • one year ago
  22. dfresenius Group Title
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    ok so here is another, im thinking with this one i can multiply by a 1/2 and use the identity sin2x

    • one year ago
  23. hartnn Group Title
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    sin x cos x = (sin 2x) /2 then we have formula for sin A cos B

    • one year ago
  24. hartnn Group Title
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    yes! good thought :)

    • one year ago
  25. dfresenius Group Title
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    nice!!

    • one year ago
  26. dfresenius Group Title
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    |dw:1361090244648:dw|

    • one year ago
  27. sami-21 Group Title
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    for the first question by parts can also be used . \[\Large \frac{1}{5}\sin^2(x)\sin(5x)-\frac{2}{5}\int\limits \sin(x)\sin(5x)\cos(x)dx\]

    • one year ago
  28. hartnn Group Title
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    sin A cos B =... ?

    • one year ago
  29. dfresenius Group Title
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    oops

    • one year ago
  30. .Sam. Group Title
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    \[( \sin A ) ( \sin B )=\frac{1}{2} (\cos (A -B )-\cos (A +B ))\]

    • one year ago
  31. .Sam. Group Title
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    But there's still a lot of work if you do this

    • one year ago
  32. dfresenius Group Title
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    |dw:1361090429038:dw|

    • one year ago
  33. dfresenius Group Title
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    |dw:1361090501054:dw|

    • one year ago
  34. hartnn Group Title
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    why - *minus* in between ?

    • one year ago
  35. dfresenius Group Title
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    integral of sin is -cosx

    • one year ago
  36. hartnn Group Title
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    both of - from that you have factored out, right ? to make -1/4 ?

    • one year ago
  37. dfresenius Group Title
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    well, before i took integral

    • one year ago
  38. dfresenius Group Title
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    |dw:1361090806081:dw|

    • one year ago
  39. dfresenius Group Title
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    those will both be negative right?

    • one year ago
  40. hartnn Group Title
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    integral of those will be negative, yes, right.

    • one year ago
  41. dfresenius Group Title
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    ok thats all i got, thank you so much both of you for helping me out, big test this friday

    • one year ago
  42. hartnn Group Title
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    best of luck (Y)

    • one year ago
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