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We're doing indefinite integrals using the substitution rule right now in class.
The problem:
(integral of) (e^6x)csc(e^6x)cot(e^6x)dx
I am calling 'u' my substitution variable. I feel like I've tried every possible substitution, but I still haven't found the right one.
The most promising substitution:
u= csc(e^6x)
so my equation would become
(integral of) usin(e^6x)
Now I don't know what to do, because we haven't learned how to the problem like this. I feel like there must be some substitution that will leave me with only one term to integrate, but I don't think I've found it. Suggestions?

  • calculus -

    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

  • calculus -

    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.

  • calculus -

    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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