How do you find the integral of #e^(2x) cos3x dx#? Calculus Techniques of Integration Integration by Parts 1 Answer maganbhai P. Mar 26, 2018 #inte^(2x)cos3xdx=e^(2x)/sqrt13 cos(2x-tan^-1(3/2)) +c# Explanation: 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# Answer link Related questions How do I find the integral #int(x*ln(x))dx# ? How do I find the integral #int(cos(x)/e^x)dx# ? How do I find the integral #int(x*cos(5x))dx# ? How do I find the integral #int(x*e^-x)dx# ? How do I find the integral #int(x^2*sin(pix))dx# ? How do I find the integral #intln(2x+1)dx# ? How do I find the integral #intsin^-1(x)dx# ? How do I find the integral #intarctan(4x)dx# ? How do I find the integral #intx^5*ln(x)dx# ? How do I find the integral #intx*2^xdx# ? See all questions in Integration by Parts Impact of this question 12878 views around the world You can reuse this answer Creative Commons License