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anonymous

  • one year ago

where do I go wrong...

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  1. anonymous
    • one year ago
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    |dw:1439117048654:dw|

  2. anonymous
    • one year ago
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    |dw:1439117111328:dw|

  3. anonymous
    • one year ago
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    the answer should be -cos6t/12+c

  4. ganeshie8
    • one year ago
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    both answers are correct

  5. hartnn
    • one year ago
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    your approach and answer is correct, you just got an equivalent answer! try to re-substitute u by sin 3t

  6. hartnn
    • one year ago
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    and then apply the double angle formula for cos 2x if you need the answer in terms of cos 6x

  7. anonymous
    • one year ago
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    in that case, finally! :D

  8. hartnn
    • one year ago
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    you got it ? you'd get sin^2 3t and would use \(\cos 2x= 1-2\sin^2 x\) formula....

  9. UsukiDoll
    • one year ago
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    I don't see anything wrong with it either. the u-sub is correct u = sin3t du = cos3t(3) dx 1/3 du = cos3t dx

  10. ganeshie8
    • one year ago
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    you could also try |dw:1439117565392:dw|

  11. hartnn
    • one year ago
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    was just about to tell that^^ :P

  12. UsukiDoll
    • one year ago
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    then for double angle formula... notice that our original problem has a 3? \[\cos 2x= 1-2\sin^2 x \] this is just standard... but by mutliplying 3 \[\cos 2(3)x= 1-2\sin^2 3x \] \[\cos 6x= 1-2\sin^2 3x \]

  13. UsukiDoll
    • one year ago
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    multiply 3 on the left side for cos2(3) x ... but the 3 is needed to fit your question on the right hand side.

  14. ganeshie8
    • one year ago
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    remember, indefinite integral gives you a whole family of functions which are unique only upto a constant \(\cos 6x = 1-2\sin^23x\) so, \(\cos 6x \equiv -2\sin^23x\pmod{\text{constant functions}}\)

  15. anonymous
    • one year ago
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    Really good tips guys, thank you all :)

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