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#