How do you find the integral of $(e^{8 x}) sin (9 x) d x$ $(e^(8x))sin(9x)dx$ ?

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How do you find the integral of $(e^{8 x}) sin (9 x) d x$ $(e^(8x))sin(9x)dx$ ?

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Answer 1 · 2018-03-07T02:22:28

$\int e^{8 x} sin (9 x) d x = \frac{1}{145} e^{8 x} (8 sin (9 x) - 9 cos (9 x)) + C$ $\inte^(8x)sin(9x) dx=1/145e^(8x)(8sin(9x)-9cos(9x)) +C$

Explanation:

(1) $\int e^{8 x} sin (9 x) d x$ $\inte^(8x)sin(9x) dx$

We need to use integration by parts.

$\int u d v = u v - \int v d u$ $\int udv = uv - \intvdu$

It's useful to remember the acronym LIATE: Log, Inverse Trig, Algebraic, Trig, Exponential for knowing the prioritization of choosing your $u$ $u$ value.

Since trigonometric functions take precedence over exponentials, we will use $u = sin (9 x)$ $u = sin(9x)$ .

Substitution:
$u = sin (9 x)$ $u= sin(9x)$
$d u = 9 cos (9 x) d x$ $du = 9cos(9x)dx$
$d v = e^{8 x} d x$ $dv = e^(8x)dx$
$v = \frac{1}{8} e^{8 x}$ $v = 1/8e^(8x)$

(2) $\int e^{8 x} sin (9 x) d x = \frac{1}{8} e^{8 x} sin (9 x) - \frac{9}{8} \int e^{8 x} cos (9 x) d x$ $\inte^(8x)sin(9x) dx = 1/8e^(8x)sin(9x)-9/8\inte^(8x)cos(9x)dx$

Since we haven't improved our situation, it looks like another round of integration by parts. Let's use the letters $y$ $y$ and $z$ $z$ , and let $y = cos (9 x)$ $y = cos(9x)$ and integrate the RHS integral by parts.

$\int y d z = y z - \int z d y$ $\int ydz = yz - \intzdy$

Substitution:
$y = cos (9 x)$ $y = cos(9x)$
$d y = - 9 sin (9 x) d x$ $dy = -9sin(9x)dx$
$d z = e^{8 x} d x$ $dz = e^(8x)dx$
$z = \frac{1}{8} e^{8 x}$ $z = 1/8e^(8x)$

(3) $\int e^{8 x} cos (9 x) d x = \frac{1}{8} e^{8 x} cos (9 x) + \frac{9}{8} \int e^{8 x} sin (9 x) d x$ $\inte^(8x)cos(9x)dx = 1/8e^(8x)cos(9x)+9/8\inte^(8x)sin(9x)dx$

Substituting (3) $\to$ $->$ (2):

(4) $\int e^{8 x} sin (9 x) d x = \frac{1}{8} e^{8 x} sin (9 x)$ $\inte^(8x)sin(9x) dx = 1/8e^(8x)sin(9x)$ $- \frac{9}{8} (\frac{1}{8} e^{8 x} cos (9 x) + \frac{9}{8} \int e^{8 x} sin (9 x) d x)$ $- 9/8 (1/8e^(8x)cos(9x)+9/8\inte^(8x)sin(9x)dx)$

$\int e^{8 x} sin (9 x) d x = \frac{1}{8} e^{8 x} sin (9 x) - \frac{9}{64} e^{8 x} cos (9 x)$ $\inte^(8x)sin(9x) dx = 1/8e^(8x)sin(9x) - 9/64 e^(8x)cos(9x)$ $- \frac{81}{64} \int e^{8 x} sin (9 x) d x$ $-81/64\inte^(8x)sin(9x)dx$

$\frac{145}{64} \int e^{8 x} sin (9 x) d x = \frac{1}{8} e^{8 x} sin (9 x) - \frac{9}{64} e^{8 x} cos (9 x)$ $145/64\inte^(8x)sin(9x) dx = 1/8e^(8x)sin(9x) - 9/64 e^(8x)cos(9x)$

$= \frac{64}{145} (\frac{1}{8} e^{8 x} sin (9 x) - \frac{9}{64} e^{8 x} cos (9 x)) + C$ $= 64/145(1/8e^(8x)sin(9x) - 9/64 e^(8x)cos(9x)) +C$

$= \frac{8}{145} e^{8 x} sin (9 x) - \frac{9}{145} e^{8 x} cos (9 x) + C$ $= 8/145e^(8x)sin(9x) - 9/145 e^(8x)cos(9x) +C$

$= \frac{1}{145} e^{8 x} (8 sin (9 x) - 9 cos (9 x)) + C$ $=1/145e^(8x)(8sin(9x)-9cos(9x)) +C$

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