# How do you find the antiderivative of e^(2x) * sin(4x)dx?

Apr 8, 2018

$I = {e}^{2 x} / 10 \left(\sin 4 x - 2 \cos 4 x\right) + c$
OR
$I = {e}^{2 x} / \sqrt{20} \sin \left(4 x - {\tan}^{-} 1 2\right) + c$

#### Explanation:

Here,

$I = \int {e}^{2 x} \sin 4 x \mathrm{dx}$

We know that,

color(red)((1)inte^(ax) sinbxdx=e^(ax)/(a^2+b^2)(asinbx- bcosbx)+c

Comparing we get, $a = 2 \mathmr{and} b = 4$

So,

$I = {e}^{2 x} / \left({2}^{2} + {4}^{2}\right) \left(2 \sin 4 x - 4 \cos 4 x\right) + c$

$I = {e}^{2 x} / 20 \left(2 \sin 4 x - 4 \cos 4 x\right) + c$

$I = {e}^{2 x} / 10 \left(\sin 4 x - 2 \cos 4 x\right) + c$

OR

We also know that,

color(red)((2)inte^(ax)sinbxdx=e^(ax)/sqrt(a^2+b^2)sin(bx- theta)+c,

where, $\theta = {\tan}^{-} 1 \left(\frac{b}{a}\right)$

So,

$I = {e}^{2 x} / \sqrt{{2}^{2} + {4}^{2}} \sin \left(4 x - \theta\right) + c$,

where,$\theta = {\tan}^{-} 1 \left(\frac{4}{2}\right) = {\tan}^{-} 1 \left(2\right)$

Hence,

$I = {e}^{2 x} / \sqrt{20} \sin \left(4 x - {\tan}^{-} 1 2\right) + c$