# How do you find the integral of  (sinx)(5^x) dx?

May 17, 2018

$\int {5}^{x} \sin x \mathrm{dx} = \frac{{5}^{x} \left(\ln 5 \sin x - \cos x\right)}{1 + {\ln}^{2} 5} + C$

#### Explanation:

Integrate by parts:

$\int {5}^{x} \sin x \mathrm{dx} = \int {5}^{x} \frac{d}{\mathrm{dx}} \left(- \cos x\right) \mathrm{dx}$

$\int {5}^{x} \sin x \mathrm{dx} = = - {5}^{x} \cos x + \int \cos x \frac{d}{\mathrm{dx}} \left({5}^{x}\right) \mathrm{dx}$

$\int {5}^{x} \sin x \mathrm{dx} = = - {5}^{x} \cos x + \ln 5 \int \cos x {5}^{x} \mathrm{dx}$

and then again:

$\int {5}^{x} \sin x \mathrm{dx} = = - {5}^{x} \cos x + \ln 5 \int \frac{d}{\mathrm{dx}} \left(\sin x\right) {5}^{x} \mathrm{dx}$

$\int {5}^{x} \sin x \mathrm{dx} = = - {5}^{x} \cos x + \ln 5 \sin x {5}^{x} - \ln 5 \int \sin x \frac{d}{\mathrm{dx}} \left({5}^{x}\right) \mathrm{dx}$

$\int {5}^{x} \sin x \mathrm{dx} = = {5}^{x} \left(\ln 5 \sin x - \cos x\right) - {\ln}^{2} 5 \int \sin x {5}^{x} \mathrm{dx}$

The integral now appears on both sides of the equation and we can solve for it:

$\left(1 + {\ln}^{2} 5\right) \int {5}^{x} \sin x \mathrm{dx} = {5}^{x} \left(\ln 5 \sin x - \cos x\right) + C$

$\int {5}^{x} \sin x \mathrm{dx} = \frac{{5}^{x} \left(\ln 5 \sin x - \cos x\right)}{1 + {\ln}^{2} 5} + C$