Question #d6918

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Question #d6918

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Answer 1 · 2017-11-25T04:34:13

$e^{4} (- \frac{1}{2} x cos (2 x) + \frac{1}{4} sin (2 x)) + C$ $e^4(-1/2xcos(2x)+1/4sin(2x))+C$

Explanation:

$\int e (e^{3} x sin (2 x) d x) = \int e^{4} x sin (2 x) d x$ $inte(e^3xsin(2x)dx)=inte^4xsin(2x)dx$

$e^{4}$ $e^4$ is a constant and can be pulled out:

$= e^{4} \int x sin (2 x) d x$ $=e^4intxsin(2x)dx$

Let's use integration by parts since we have a product of two functions:

$u = x$ $u=x$ and $d v = sin (2 x) d x$ $dv=sin(2x)dx$ . Then:

$d u = d x$ $du=dx$ and $v = - \frac{1}{2} cos (2 x)$ $v=-1/2cos(2x)$

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

$\int x sin (2 x) d x = - \frac{1}{2} x cos (2 x) - \int - \frac{1}{2} cos (2 x) d x = - \frac{1}{2} x cos (2 x) + \frac{1}{2} \int cos (2 x) d x$ $intxsin(2x)dx=-1/2xcos(2x)-int-1/2cos(2x)dx=-1/2xcos(2x)+1/2intcos(2x)dx$

$= - \frac{1}{2} x cos (2 x) + \frac{1}{2} \cdot \frac{1}{2} sin (2 x) + C$ $=-1/2xcos(2x)+1/2*1/2sin(2x)+C$

Now we need to make sure we do not forget about the $e^{4}$ $e^4$ :

$e^{4} \int x sin (2 x) d x = e^{4} (- \frac{1}{2} x cos (2 x) + \frac{1}{4} sin (2 x)) + C$ $e^4intxsin(2x)dx=e^4(-1/2xcos(2x)+1/4sin(2x))+C$

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Question #d6918

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