# What is int \ e^(5x+1) sin(2x+3) \ dx ?

Sep 9, 2017

$\int \setminus {e}^{5 x + 1} \sin \left(2 x + 3\right) \setminus \mathrm{dx} = {e}^{5 x + 1} \setminus \frac{\left(5 \sin \left(2 x + 3\right) - 2 \cos \left(2 x + 3\right)\right)}{29} + C$

#### Explanation:

We seek:

$I = \int \setminus {e}^{5 x + 1} \sin \left(2 x + 3\right) \setminus \mathrm{dx}$

Utilising the given result:

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

We now perform a substitution:

Let $u = 2 x + 3 \implies \frac{\mathrm{du}}{\mathrm{dx}} = 2$; and $x = \frac{1}{2} \left(u - 3\right)$

Substituting into the integral we get:

$I = \int \setminus {e}^{\frac{5}{2} \left(u - 3\right) + 1} \sin u \setminus \left(\frac{1}{2}\right) \setminus \mathrm{du}$
$\setminus \setminus = \frac{1}{2} \setminus \int \setminus {e}^{\frac{5}{2} u - \frac{15}{2} + 1} \sin u \setminus \mathrm{du}$
$\setminus \setminus = \frac{1}{2} \setminus \int \setminus {e}^{\frac{5}{2} u} {e}^{- \frac{13}{2}} \sin u \setminus \mathrm{du}$
$\setminus \setminus = \frac{1}{2} {e}^{- \frac{13}{2}} \int \setminus {e}^{\frac{5}{2} u} \sin u \setminus \mathrm{du}$

We can now use the given result [A] with:

$a = \frac{5}{2}$
$b = 1$

Giving:

$I = \frac{1}{2} {e}^{- \frac{13}{2}} \left\{\frac{{e}^{\frac{5}{2} u} \left(\frac{5}{2} \sin 1 u - 1 \cos 1 u\right)}{{\left(\frac{5}{2}\right)}^{2} + {1}^{2}}\right\} + C$
$\setminus \setminus = \frac{1}{2} {e}^{- \frac{13}{2}} \left\{\frac{{e}^{\frac{5}{2} u} \setminus \frac{1}{2} \setminus \left(5 \sin u - 2 \cos u\right)}{\frac{25}{4} + 1}\right\} + C$
$\setminus \setminus = \frac{1}{4} {e}^{- \frac{13}{2}} \left\{\frac{{e}^{\frac{5}{2} u} \setminus \left(5 \sin u - 2 \cos u\right)}{\frac{29}{4}}\right\} + C$
$\setminus \setminus = {e}^{- \frac{13}{2}} \cdot \frac{{e}^{\frac{5}{2} u} \setminus \left(5 \sin u - 2 \cos u\right)}{29} + C$
$\setminus \setminus = \frac{{e}^{\frac{5}{2} u - \frac{13}{2}} \setminus \left(5 \sin u - 2 \cos u\right)}{29} + C$

Now if we restore the earlier substitution:

$I = \frac{{e}^{\frac{5}{2} \left(2 x + 3\right) - \frac{13}{2}} \setminus \left(5 \sin \left(2 x + 3\right) - 2 \cos \left(2 x + 3\right)\right)}{29} + C$
$\setminus \setminus = \frac{{e}^{5 x + \frac{15}{2} - \frac{13}{2}} \setminus \left(5 \sin \left(2 x + 3\right) - 2 \cos \left(2 x + 3\right)\right)}{29} + C$
$\setminus \setminus = \frac{{e}^{5 x + 1} \setminus \left(5 \sin \left(2 x + 3\right) - 2 \cos \left(2 x + 3\right)\right)}{29} + C$