How do you integrate $\int e^{2 x} sin (3 x)$ $int e^(2x)sin(3x)$ by integration by parts method?

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How do you integrate $\int e^{2 x} sin (3 x)$ $int e^(2x)sin(3x)$ by integration by parts method?

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Answer 1 · 2017-12-28T01:44:45

$\int e^{2 x} sin (3 x) d x = - \frac{3}{13} e^{2 x} cos (3 x) + \frac{2}{13} e^{2 x} sin (3 x)$ $inte^(2x)sin(3x)dx=-3/13e^(2x)cos(3x)+2/13e^(2x)sin(3x)$

Explanation:

$\int e^{2 x} sin (3 x) d x$ $inte^(2x)sin(3x)dx$

if $u = e^{2 x}$ $u=e^(2x)$ and $d v = sin (3 x) d x$ $dv=sin(3x)dx$ , then $d u = 2 e^{2 x} d x$ $du=2e^(2x)dx$ and $v = - \frac{1}{3} cos (3 x)$ $v=-1/3cos(3x)$ .

using integration by parts, the integral becomes:

$- \frac{1}{3} e^{2 x} cos (3 x) - \int - \frac{1}{3} \cdot 2 e^{2 x} cos (3 x) d x$ $-1/3e^(2x)cos(3x)-int-1/3*2e^(2x)cos(3x)dx$
$= - \frac{1}{3} e^{2 x} cos (3 x) + \frac{2}{3} \int e^{2 x} cos (3 x) d x$ $=-1/3e^(2x)cos(3x)+2/3inte^(2x)cos(3x)dx$

integrate by parts again:

$u = e^{2 x}$ $u=e^(2x)$ and $d v = cos (3 x) d x$ $dv=cos(3x)dx$ , which means $d u = 2 e^{2 x} d x$ $du=2e^(2x)dx$ and $v = \frac{1}{3} sin (3 x)$ $v=1/3sin(3x)$

the integral is now:
$= - \frac{1}{3} e^{2 x} cos (3 x) + \frac{2}{3} (\frac{1}{3} e^{2 x} sin (3 x) - \int 2 \cdot \frac{1}{3} e^{2 x} sin (3 x) d x)$ $=-1/3e^(2x)cos(3x)+2/3(1/3e^(2x)sin(3x)-int2*1/3e^(2x)sin(3x)dx)$
$= - \frac{1}{3} e^{2 x} cos (3 x) + \frac{2}{9} e^{2 x} sin (3 x) - \frac{2}{3} \int \frac{2}{3} e^{2 x} sin (3 x) d x$ $=-1/3e^(2x)cos(3x)+2/9e^(2x)sin(3x)-2/3int2/3e^(2x)sin(3x)dx$
$= - \frac{1}{3} e^{2 x} cos (3 x) + \frac{2}{9} e^{2 x} sin (3 x) - \frac{4}{9} \int e^{2 x} sin (3 x) d x$ $=-1/3e^(2x)cos(3x)+2/9e^(2x)sin(3x)-4/9inte^(2x)sin(3x)dx$

remember that this expression is equal to the original integral, meaning:

$\int e^{2 x} sin (3 x) d x = - \frac{1}{3} e^{2 x} cos (3 x) + \frac{2}{9} e^{2 x} sin (3 x) - \frac{4}{9} \int e^{2 x} sin (3 x) d x$ $inte^(2x)sin(3x)dx=-1/3e^(2x)cos(3x)+2/9e^(2x)sin(3x)-4/9inte^(2x)sin(3x)dx$

solve for $\int e^{2 x} sin (3 x) d x$ $inte^(2x)sin(3x)dx$ :

$\frac{13}{9} \int e^{2 x} sin (3 x) d x = - \frac{1}{3} e^{2 x} cos (3 x) + \frac{2}{9} e^{2 x} sin (3 x)$ $13/9inte^(2x)sin(3x)dx=-1/3e^(2x)cos(3x)+2/9e^(2x)sin(3x)$
$\int e^{2 x} sin (3 x) d x = \frac{9}{13} (- \frac{1}{3} e^{2 x} cos (3 x) + \frac{2}{9} e^{2 x} sin (3 x))$ $inte^(2x)sin(3x)dx=9/13(-1/3e^(2x)cos(3x)+2/9e^(2x)sin(3x))$
$\int e^{2 x} sin (3 x) d x = - \frac{3}{13} e^{2 x} cos (3 x) + \frac{2}{13} e^{2 x} sin (3 x)$ $inte^(2x)sin(3x)dx=-3/13e^(2x)cos(3x)+2/13e^(2x)sin(3x)$

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How do you integrate $\int e^{2 x} sin (3 x)$ $int e^(2x)sin(3x)$ by integration by parts method?

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