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

Question

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

Answer 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

Answer 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!

Answer 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.