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dfresenius

  • 3 years ago

Integral with trig functions

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  1. dfresenius
    • 3 years ago
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    |dw:1361089150658:dw|

  2. dumbcow
    • 3 years ago
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    i am intrigued

  3. dfresenius
    • 3 years ago
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    |dw:1361089181888:dw|

  4. dfresenius
    • 3 years ago
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    or would it be cos^xcos5x

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

  6. dfresenius
    • 3 years ago
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    ok

  7. dfresenius
    • 3 years ago
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    so left integral is easy

  8. dfresenius
    • 3 years ago
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    hmm no i cant do that

  9. dfresenius
    • 3 years ago
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    that 5x is just blowing my mind

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

  11. hartnn
    • 3 years ago
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    huh ? integral of cos 5x is just sin 5x/5 +c .

  12. dfresenius
    • 3 years ago
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    |dw:1361089595564:dw|

  13. dfresenius
    • 3 years ago
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    yA i know that integral, but its the cos^2 that is messin me up

  14. hartnn
    • 3 years ago
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    ok, you need to used \(\sin^2x = (1- \cos2x )/2\)

  15. dfresenius
    • 3 years ago
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    okkk

  16. hartnn
    • 3 years ago
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    then, for cos 2x cos 7x there is a standard formula for , cos A cos B =...

  17. hartnn
    • 3 years ago
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    *cos 2x cos 5x

  18. dumbcow
    • 3 years ago
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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

  19. dfresenius
    • 3 years ago
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    i got it then :)

  20. hartnn
    • 3 years ago
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    good!

  21. dfresenius
    • 3 years ago
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    |dw:1361090139384:dw|

  22. dfresenius
    • 3 years ago
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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

  23. hartnn
    • 3 years ago
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    sin x cos x = (sin 2x) /2 then we have formula for sin A cos B

  24. hartnn
    • 3 years ago
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    yes! good thought :)

  25. dfresenius
    • 3 years ago
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    nice!!

  26. dfresenius
    • 3 years ago
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    |dw:1361090244648:dw|

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

  28. hartnn
    • 3 years ago
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    sin A cos B =... ?

  29. dfresenius
    • 3 years ago
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    oops

  30. .Sam.
    • 3 years ago
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    \[( \sin A ) ( \sin B )=\frac{1}{2} (\cos (A -B )-\cos (A +B ))\]

  31. .Sam.
    • 3 years ago
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    But there's still a lot of work if you do this

  32. dfresenius
    • 3 years ago
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    |dw:1361090429038:dw|

  33. dfresenius
    • 3 years ago
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    |dw:1361090501054:dw|

  34. hartnn
    • 3 years ago
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    why - *minus* in between ?

  35. dfresenius
    • 3 years ago
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    integral of sin is -cosx

  36. hartnn
    • 3 years ago
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    both of - from that you have factored out, right ? to make -1/4 ?

  37. dfresenius
    • 3 years ago
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    well, before i took integral

  38. dfresenius
    • 3 years ago
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    |dw:1361090806081:dw|

  39. dfresenius
    • 3 years ago
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    those will both be negative right?

  40. hartnn
    • 3 years ago
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    integral of those will be negative, yes, right.

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

  42. hartnn
    • 3 years ago
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    best of luck (Y)

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