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

Jan 20, 2017

$- \frac{1}{\ln} 4 \cos \left({4}^{x}\right) + C$

#### 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.

$\ln u = \ln \left({4}^{x}\right)$

$\frac{1}{u} \left(\mathrm{du}\right) = \ln 4 \mathrm{dx}$

$\mathrm{du} = u \ln 4 \mathrm{dx}$

$\mathrm{du} = {4}^{x} \ln 4 \mathrm{dx}$

$\frac{\mathrm{du}}{{4}^{x} \ln 4} = \mathrm{dx}$

Now substitute:

$\int {4}^{x} \sin u \cdot \frac{\mathrm{du}}{{4}^{x} \ln 4}$

$\frac{1}{\ln} 4 \int \sin u \mathrm{du}$

$- \frac{1}{\ln} 4 \cos u + C$

$- \frac{1}{\ln} 4 \cos \left({4}^{x}\right) + C$

Hopefully this helps!