How do you find the integral of $e^(2x) cos3x dx$ ?

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Answer 1 · 2018-03-26T10:26:17

$inte^(2x)cos3xdx=e^(2x)/sqrt13 cos(2x-tan^-1(3/2)) +c$

We know that,

$color(red)(inte^(ax)cosbx dx=e^(ax)/(sqrt(a^2+b^2))cos(bx-theta)+c$

where, $color(red)(costheta=a/(sqrt(a^2+b^2)) and sintheta=b/(sqrt(a^2+b^2))$

Substituting , $a=2,and b=3$ , we get

$costheta=2/sqrt(2^2+3^2)=2/sqrt13 and sintheta=3/sqrt(2^2+3^2) =3/sqrt13$

$=>tantheta=sintheta/costheta = (3/sqrt13)/(2/(sqrt13))=3/2=>theta=tan^-1(3/2)$

So,

$inte^(2x)cos3xdx=e^(2x)/sqrt13 cos(2x-tan^-1(3/2)) +c$

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