# How do you integrate int e^x cos x dx  using integration by parts?

Jun 22, 2016

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

#### Explanation:

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

$= \setminus m a t h c a l \left\{R e\right\} \left(\int \setminus {e}^{x} \left(\cos x + i \setminus \sin x\right) \setminus \mathrm{dx}\right)$

$= \setminus m a t h c a l \left\{R e\right\} \left(\int \setminus {e}^{x} {e}^{i x} \setminus \mathrm{dx}\right)$

$= \setminus m a t h c a l \left\{R e\right\} \left(\int \setminus {e}^{\left(1 + i\right) x} \setminus \mathrm{dx}\right)$

$= \setminus m a t h c a l \left\{R e\right\} \left(\frac{1}{1 + i} {e}^{\left(1 + i\right) x}\right)$

$= {e}^{x} \setminus m a t h c a l \left\{R e\right\} \left(\frac{1}{1 + i} {e}^{i x}\right)$

$= {e}^{x} \setminus m a t h c a l \left\{R e\right\} \left(\frac{1 - i}{\left(1 + i\right) \left(1 - i\right)} \left(\cos x + i \sin x\right)\right)$

$= {e}^{x} \setminus m a t h c a l \left\{R e\right\} \left(\frac{1 - i}{2} \left(\cos x + i \sin x\right)\right)$

$= \frac{1}{2} {e}^{x} \setminus m a t h c a l \left\{R e\right\} \left(\left(1 - i\right) \left(\cos x + i \sin x\right)\right)$

$= \frac{1}{2} {e}^{x} \setminus m a t h c a l \left\{R e\right\} \left(\cos x + i \sin x - i \cos x + \sin x\right)$

$= \frac{1}{2} {e}^{x} \setminus m a t h c a l \left\{R e\right\} \left(\left(\cos x + \sin x\right) + i \left(\sin x - \cos x\right)\right)$

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