anonymous
  • anonymous
where do I go wrong...
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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anonymous
  • anonymous
|dw:1439117048654:dw|
anonymous
  • anonymous
|dw:1439117111328:dw|
anonymous
  • anonymous
the answer should be -cos6t/12+c

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More answers

ganeshie8
  • ganeshie8
both answers are correct
hartnn
  • hartnn
your approach and answer is correct, you just got an equivalent answer! try to re-substitute u by sin 3t
hartnn
  • hartnn
and then apply the double angle formula for cos 2x if you need the answer in terms of cos 6x
anonymous
  • anonymous
in that case, finally! :D
hartnn
  • hartnn
you got it ? you'd get sin^2 3t and would use \(\cos 2x= 1-2\sin^2 x\) formula....
UsukiDoll
  • UsukiDoll
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
ganeshie8
  • ganeshie8
you could also try |dw:1439117565392:dw|
hartnn
  • hartnn
was just about to tell that^^ :P
UsukiDoll
  • UsukiDoll
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 \[\cos 2(3)x= 1-2\sin^2 3x \] \[\cos 6x= 1-2\sin^2 3x \]
UsukiDoll
  • UsukiDoll
multiply 3 on the left side for cos2(3) x ... but the 3 is needed to fit your question on the right hand side.
ganeshie8
  • ganeshie8
remember, indefinite integral gives you a whole family of functions which are unique only upto a constant \(\cos 6x = 1-2\sin^23x\) so, \(\cos 6x \equiv -2\sin^23x\pmod{\text{constant functions}}\)
anonymous
  • anonymous
Really good tips guys, thank you all :)

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