Question:

The integral of 4e^(4x)sin(e^(4x))?

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I have tried very hard and my calc book is not helpful...

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  1. Let u = e^(4x).  Then du = 4e^(4x) dx.  So:

    ∫ 4e^(4x) sin(e^(4x)) dx =

    ∫ sin(u) du =

    -cos(u) + C =

    -cos(e^(4x)) + C


  2. ∫ 4e^(4x) sin[e^(4x)] dx =

    the first thing you should notice is that (fortunately) your integrand includes

    both the function e^(4x) (the sine argument) and its derivative, that is just 4e^(4x);

    thus substitute e^(4x) = u

    differentiate both sides into:

    4e^(4x) dx = du

    then, substituting, you get:

    ∫  sin[e^(4x)] 4e^(4x) dx = ∫ sin u du = - cos u + C

    thus, being u = e^(4x), you finally get:

    ∫ 4e^(4x) sin[e^(4x)] dx = - cos[e^(4x)] + C

    I hope it helps..

    Bye!


  3. Let u = e^(4x).  Then du = 4e^(4x) dx, so we can rewrite the integral as:

    Integral(sin(u) du) = -cos(u) + C = -cos(e^(4x)) + C.

    You can check that this is correct by differentiating -cos(e^(4x)) + C.

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