How do you integrate $int e^(5x)cos3x$ ?

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Answer 1 · 2017-09-19T02:08:29

You can integrate by parts twice and algebraically solve for the integral to get $int\ e^{5x}cos 3x\ dx=5/34 e^{5x}cos 3x+3/34 e^{5x}sin 3x+C$

Let $I=int\ e^{5x}cos 3x\ dx$ . Now let $u=e^{5x}$ and $dv=cos 3x\ dx$ so that $du=5e^{5x}\ dx$ and $v=1/3 sin 3x$ .

Then $I=1/3 e^{5x}sin 3x-5/3int\ e^{5x}sin 3x\ dx$ .

For this next integral, let $u=e^{5x}$ and $dv=sin 3x\ dx$ so that $du=5e^{5x}$ and $v=-1/3 cos 3x$ . It follows that

$I=1/3 e^{5x}sin 3x-5/3(-1/3 e^{5x}cos 3x+5/3 int e^{5x}cos 3x\ dx)$

$=1/3 e^{5x}sin 3x+5/9e^{5x}cos 3x-25/9 I$ .

Therefore, $34/9 I=1/3 e^{5x}sin 3x+5/9 e^{5x}cos 3x$ .

Multiplying both sides by $9/34$ , rearranging, and tacking on a $+C$ at the end gives

$I=int\ e^{5x}cos 3x\ dx=5/34 e^{5x}cos 3x+3/34 e^{5x}sin 3x+C$ .

How do you integrate int e^(5x)cos3xint e^(5x)cos3x?