Evaluate the integral? int e^(2x) cosx dx

Jul 5, 2017

See below.

Explanation:

Using the de Moivre's identity

${e}^{i x} = \cos x + i \sin x$ we have

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

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

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

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

Taking the real part of this integral we have

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

I have solved with another way:

Jul 5, 2017

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

Explanation:

Let:

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

We can use integration by parts:

Let $\left\{\begin{matrix}u & = \cos x & \implies \frac{\mathrm{du}}{\mathrm{dx}} & = - \sin x \\ \frac{\mathrm{dv}}{\mathrm{dx}} & = {e}^{2 x} & \implies v & = \frac{1}{2} {e}^{2 x}\end{matrix}\right.$

Then plugging into the IBP formula:

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

gives us

$\int \setminus \left(\cos x\right) \left({e}^{2 x}\right) \setminus \mathrm{dx} = \left(\cos x\right) \left(\frac{1}{2} {e}^{2 x}\right) - \int \setminus \left(\frac{1}{2} {e}^{2 x}\right) \left(- \sin x\right) \setminus \mathrm{dx}$
$\therefore I = \frac{1}{2} {e}^{2 x} \cos x + \frac{1}{2} \int \setminus {e}^{2 x} \setminus \sin x \setminus \mathrm{dx}$ .... [A]

At first it appears as if we have made no progress, as now the second integral is similar to $I$, having exchanged $\cos x$ for $\sin x$, but if we apply IBP a second time then the progress will become clear:

Let $\left\{\begin{matrix}u & = \sin x & \implies \frac{\mathrm{du}}{\mathrm{dx}} & = \cos x \\ \frac{\mathrm{dv}}{\mathrm{dx}} & = {e}^{2 x} & \implies v & = \frac{1}{2} {e}^{2 x}\end{matrix}\right.$

Then plugging into the IBP formula, gives us:

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

Inserting this result into [A] we get:

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

$\therefore I = \frac{1}{2} {e}^{2 x} \cos x + \frac{1}{4} {e}^{2 x} \sin x - \frac{1}{4} I + A$

$\therefore 4 I = 2 {e}^{2 x} \cos x + {e}^{2 x} \sin x - I + 4 A$

$\therefore 5 I = 2 {e}^{2 x} \cos x + {e}^{2 x} \sin x + 4 A$

$\therefore I = \frac{2}{5} {e}^{2 x} \cos x + \frac{1}{5} {e}^{2 x} \sin x + C$