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

Question

1335 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-01-20T02:39:30

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

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!

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