# How do you use substitution to integrate e^x * sin2xdx ?

May 14, 2018

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

#### Explanation:

We seek the integral:

$I = \int \setminus {e}^{x} \sin 2 x \setminus \mathrm{dx}$

There is no suitable substitution, however, We can apply Integration By Parts:

Let $\left\{\begin{matrix}u & = \sin 2 x & \implies \frac{\mathrm{du}}{\mathrm{dx}} & = 2 \cos 2 x \\ \frac{\mathrm{dv}}{\mathrm{dx}} & = {e}^{x} & \implies v & = {e}^{x}\end{matrix}\right.$

Then plugging into the IBP formula:

$\int \setminus \left(u\right) \left(\frac{\mathrm{dv}}{\mathrm{dx}}\right) \setminus \mathrm{dx} = \left(u\right) \left(v\right) - \int \setminus \left(v\right) \left(\frac{\mathrm{du}}{\mathrm{dx}}\right) \setminus \mathrm{dx}$

We have:

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

$\therefore I = {e}^{2 x} \sin 2 x - 2 \setminus \int \setminus {e}^{x} \cos 2 x \setminus \mathrm{dx}$

Now consider the integral given by:

${I}_{2} = \int \setminus {e}^{x} \cos 2 x \setminus \mathrm{dx}$

We will now need to apply IBP again:

Let $\left\{\begin{matrix}u & = \cos 2 x & \implies \frac{\mathrm{du}}{\mathrm{dx}} & = - 2 \sin 2 x \\ \frac{\mathrm{dv}}{\mathrm{dx}} & = {e}^{x} & \implies v & = {e}^{x}\end{matrix}\right.$

Then plugging into the IBP formula we have::

$\int \setminus \left(\cos 2 x\right) \left({e}^{x}\right) \setminus \mathrm{dx} = \left(\cos 2 x\right) \left({e}^{x}\right) - \int \setminus \left({e}^{x}\right) \left(- 2 \sin 2 x\right) \setminus \mathrm{dt}$

${I}_{2} = {e}^{x} \cos 2 x + 2 \setminus \int \setminus {e}^{x} \sin 2 x \setminus \mathrm{dt}$
$\setminus \setminus \setminus = {e}^{x} \cos 2 x + 2 I$

And so combining the results we find that:

$I = {e}^{x} \sin 2 x - 2 \left\{{e}^{x} \cos 2 x + 2 I\right\}$
$\setminus \setminus = {e}^{x} \sin 2 x - 2 {e}^{x} \cos 2 x - 4 I$

$\therefore 5 I = {e}^{x} \left(\sin 2 x - 2 \cos 2 x\right)$

$\therefore I = \frac{1}{5} {e}^{x} \left(\sin 2 x - 2 \cos 2 x\right)$

And not forgetting the constant of integration,

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