# How do you integrate int e^6xcos5xdx?

Mar 7, 2015

Hello !

I propose another solution, shorter (?), using complex numbers.

${e}^{6 x} \cos \left(5 x\right) = {e}^{6 x} \setminus m a t h \mathfrak{a} k \left\{R e\right\} \left({e}^{5 i x}\right) = \setminus m a t h \mathfrak{a} k \left\{R e\right\} \left({e}^{\left(6 + 5 i\right) x}\right)$.

Now integrate easily :

$\int {e}^{6 x} \cos \left(5 x\right) \mathrm{dx} = \setminus m a t h \mathfrak{a} k \left\{R e\right\} \setminus \int {e}^{\left(6 + 5 i\right) x} d x = \setminus m a t h \mathfrak{a} k \left\{R e\right\} \frac{{e}^{\left(6 + 5 i\right) x}}{6 + 5 i} + c$

where $c \setminus \in \mathbb{C}$ is a constant.

To take the real part, you have to write :

$\frac{{e}^{\left(6 + 5 i\right) x}}{6 + 5 i} = {e}^{6 x} \frac{\left(\cos \left(5 x\right) + i \sin \left(5 x\right)\right) \left(6 - 5 i\right)}{\left(6 + 5 i\right) \left(6 - 5 i\right)} = {e}^{6 x} \frac{\left(\cos \left(5 x\right) + i \sin \left(5 x\right)\right) \left(6 - 5 i\right)}{36 + 25}$

so,
$\setminus m a t h \mathfrak{a} k \left\{R e\right\} \frac{{e}^{\left(6 + 5 i\right) x}}{6 + 5 i} = {e}^{6 x} \frac{6 \cos \left(5 x\right) + 5 \sin \left(5 x\right)}{61}$.

Finally,

$\int {e}^{6 x} \cos \left(5 x\right) \mathrm{dx} = {e}^{6 x} \left(\frac{6}{61} \cos \left(5 x\right) + \frac{5}{61} \sin \left(5 x\right)\right) + c$