How do you find the integral of $f (x) = e^{2 x} sin 3 x$ $f(x)=e^(2x)sin3x$ using integration by parts?

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How do you find the integral of $f (x) = e^{2 x} sin 3 x$ $f(x)=e^(2x)sin3x$ using integration by parts?

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Answer 1 · 2015-11-05T18:32:29

$\frac{1}{13} e^{2 x} (2 sin (3 x) - 3 cos (3 x)) + c$ $1/13e^(2x)(2sin(3x) - 3cos(3x)) + c$

Explanation:

So, with integration by parts we state that:

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

Let's use $e^{2 x}$ $e^(2x)$ as our u function and $sin 3 x$ $sin3x$ as our v' function, leaving us with:

$u = e^{2 x}$ $u = e^(2x)$
$u' = 2e^(2x)$

$v = -1/3cos3x$
$v' = sin3x$

Now let's plug this back into our original formula:

$e^(2x)(-1/3cos3x) - int2e^(2x)(-1/3cos3x)dx$

Let's simplify it a bit more:

$-1/3e^(2x)(cos3x) - (-2/3)inte^(2x)(cos3x)dx$

We have to apply integration by parts yet again, using:

$u = e^(2x)$
$u' = 2e^(2x)$

$v'=cos(3x)$
$v=1/3sin(3x)$

Ultimately giving us:

$-1/3e^(2x)cos(3x)-(-2/3(e^(2x)1/3sin(3x)-int2e^(2x)1/3sin(3x)dx))$

Now, if we isolate $intsin(3x)e^(2x)dx$ from this equation we'll end up with:

$intsin(3x)e^(2x)dx = 1/13e^(2x)(2sin(3x)-3cos(3x)) + C$

I'm taking Calc 2 as well, so if anyone has a correction please don't hesitate! Also I'm running late for class so I kinda glossed over the last step, I can explain further if needed!

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How do you find the integral of $f (x) = e^{2 x} sin 3 x$ $f(x)=e^(2x)sin3x$ using integration by parts?

1 Answer

Explanation:

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How do you find the integral of $f (x) = e^{2 x} sin 3 x$ $f(x)=e^(2x)sin3x$ using integration by parts?