# How to integrate (sinx)(cosx)(e^cosx)dx ?

Apr 15, 2018

$- \cos x {e}^{\cos} x - {e}^{\cos} x + C$

#### Explanation:

Apply substitution

$\cos x = z$
$- \sin x \mathrm{dx} = \mathrm{dz}$

$\int \sin x \cos x {e}^{\cos} x = \int - z {e}^{z} \mathrm{dz}$

Apply Integration by Parts

$\int u \mathrm{dv} = u v - \int v \mathrm{du}$

$- z = u$$\textcolor{b l u e}{\rightarrow}$$- \mathrm{dz} = \mathrm{du}$

${e}^{z} \mathrm{dz} = \mathrm{dv}$$\textcolor{b l u e}{\rightarrow}$${e}^{z} = v$

$\int - z {e}^{z} \mathrm{dz}$=$- z {e}^{z} - \int - {e}^{z} \mathrm{dz}$

=$- z {e}^{z} - {e}^{z} + C$

Reverse the substitution

$\int \sin x \cos x {e}^{\cos} x = = - \cos x {e}^{\cos} x - {e}^{\cos} x + C$

Apr 15, 2018

$= - \setminus {e}^{\cos x} \left(\cos x - 1\right) + C$

#### Explanation:

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

By IBP:

$= - \int d \left({e}^{\cos} x\right) \setminus \cos x$

$= - \left(\cos x \setminus {e}^{\cos x} - \int {e}^{\cos} x d \left(\cos x\right)\right)$

Remember that: $\int {e}^{Q} \setminus d Q = {e}^{Q} + C$, so we have:

$= - \left(\cos x \setminus {e}^{\cos x} - {e}^{\cos} x + C\right)$

$= - \setminus {e}^{\cos x} \left(\cos x - 1\right) + C$