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

1 Answer
Noah G
Jan 20, 2017

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

Explanation:

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!

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