Posted by Joshua on Saturday, May 2, 2009 at 4:14pm.
you are so close
now let's do some reverse "thinking"
it looks like you know that if
y = cscx, the dy/dx = -cscx cotx
now what about
y = csc(e^6x) ?
wouldn't dy/dx = 6e^(6x)(-csc(e^6x))(cot(e^6x))
= -6e^(6x)(csc(e^6x))(cot(e^6x)) ?
Now compare that with what was given.
the only "extra" is see is the -6 in front, and that is merely a constant, so let's fudge it.
then (integral of) (e^6x)csc(e^6x)cot(e^6x)dx
= -(1/6)csc(e^6x) + C
I know that your answer is right because it is one of the options on my homework sheet, but I don't think I quite understand how everything cancels out.
If we call csc(e^6x) y from the beginning, upon initial substitution we have:
(integral of) (e^6x)(y)(cot(e^6x))dx
now dy/dx= -6e^(6x)csc(e^6x)cot(e^6x)
However, in our integral equation we have dx, not dy/dx, so we need to rearrange this so that we can directly substitute for dx
which would give us dy/(--6e^(6x)csc(e^6x)cot(e^6x)) = dx.
If you plug that into our integral, we have
-1/6 (integral of) y * 1/csc(e^6x) dy
or -1/6 (integral of) sin(e^6x)y dy
I know this isn't right, so I feel like I don't properly understand what to do with the dy/dx situation.
Never mind, I figured it out; because y= csc(e^6x), the ys cancel out, so we are left with -1/6 (integral of) dy, which gives us -1/6 csc(e^6x) + C. Thank you for the help!
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