How do you evaluate the integral #int 4^xsin(4^x)#?

Question

Emma Aldridge · Accepted Answer

#-1/ln4cos(4^x) + C#

Because there are two #4^x#'s in the integral, your best hope would probably to do a u-substitution, letting #u = 4^x#.

Differentiate using logarithmic differentiation.

#lnu = ln(4^x)#

#1/u(du) = ln4dx#

#du = u ln4dx#

#du = 4^xln4dx#

#(du)/(4^xln4) = dx#

Now substitute:

#int4^xsinu * (du)/(4^xln4)#

#1/ln4intsinudu#

#-1/ln4cosu + C#

#-1/ln4cos(4^x) + C#

Hopefully this helps!

Adam Adkins · Answer

undefined To evaluate the integral $ \int 4^x \sin(4^x) $, we use the substitution method.

Let $ u = 4^x $, then $ du = (4^x)'\, dx = \ln(4) \cdot 4^x \, dx $.

We see that $ \ln(4) \cdot 4^x \, dx = du $, and we have $ \sin(u) \, du $.

So, the integral becomes $ \int \sin(u) \, du $.

This integral is straightforward to solve, yielding $ -\cos(u) + C $, where $ C $ is the constant of integration.

Substituting back $ u = 4^x $, we have $ -\cos(4^x) + C $ as the antiderivative.

Therefore, $ \int 4^x \sin(4^x) \, dx = -\cos(4^x) + C $.