# How do you integrate sin(x)*(e)^(2 x) dx?

Jul 1, 2016

Use integration by parts (IBP).

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

#### Explanation:

To obtain the formula for IBP, we start with the product rule:

$\left(u \cdot v\right) ' = u ' \cdot v + u \cdot v '$ where u and v are functions of x in this case

Rearranging:

$u \cdot v ' = \left(u \cdot v\right) ' - u ' \cdot v$

Integrating:

$\int u \cdot v ' \mathrm{dx} = u \cdot v - \int u ' \cdot v \mathrm{dx}$

For this integral, we are going to set $u \left(x\right) = \sin \left(x\right)$ and $v ' \left(x\right) = {e}^{2 x}$

$u ' \left(x\right) = \cos \left(x\right)$ and $v \left(x\right) = \frac{1}{2} {e}^{2 x}$

So, $\int {e}^{2 x} \sin x \mathrm{dx} = \frac{1}{2} {e}^{2 x} \sin \left(x\right) - \frac{1}{2} \cdot \int \cos \left(x\right) {e}^{2 x} \mathrm{dx}$

Now do IBP again on the second term:

$u \left(x\right) = \cos \left(x\right)$ and $v ' \left(x\right) = {e}^{2 x}$
$u ' \left(x\right) = - \sin \left(x\right)$ and $v \left(x\right) = \frac{1}{2} {e}^{2 x}$

$\int {e}^{2 x} \sin \left(x\right) \mathrm{dx} = \frac{1}{2} {e}^{2 x} \sin \left(x\right) - \frac{1}{4} {e}^{2 x} \cos \left(x\right) - \frac{1}{4} \cdot \int {e}^{2 x} \sin \left(x\right) \mathrm{dx}$

As you can see, the integral on the LHS is the same as the integral on the RHS so we collect like terms and obtain:

$\frac{5}{4} \cdot \int {e}^{2 x} \sin \left(x\right) \mathrm{dx} = \frac{1}{2} {e}^{2 x} \sin \left(x\right) - \frac{1}{4} {e}^{2 x} \cos \left(x\right)$

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