Question

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

Answer 1

`-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!