Question

Using integration by Parts: Integrate e^(2x)sin(3x)dx. I can't seem to pick the right u and dv. I've trued u= e^2x and dv = sin3x dx. then tried it the other way around

Answer 1

That's e^(2x) * sin(3x) ... dx

Answer 2

Google "Pauls Online Math Notes". Then start reading and using example. Good Luck!

Answer 3

\[\int\limits_{}f(x)g(x)dx= F(x)g(x)-\int\limits_{}F(x)g'(x)dx\]
Where F(x) in the anitderivative of f(x)
\[\int\limits_{}e ^{2x}\sin(3x)dx\] Use integration by parts twice and you will notice a pattern
\[f(x)= e ^{2x}, F(x)=e ^{2x}/2, g(x)=\sin(3x), g'(x)=3\cos(3x)\]
\[1/2(e ^{2x}\sin(3x))-3/2\int\limits_{}e ^{2x}\cos(3x)dx\]
Integrate by parts once more
\[f(x)= e ^{2x}, F(x)=e ^{2x}/2, g(x)=\cos(3x), g'(x)=-3\sin(3x)\]
\[1/2(e ^{2x}\sin(3x))-3/2\left[ 1/2(e ^{2x}\cos(3x))+3/2\int\limits_{}e ^{2x}\sin(3x)dx \right]\]
Notice how this integral looks exactly like the one you started with, apart from the (-9/4) factor, all you do now is equate your original integral with this equation
\[\int\limits\limits_{}e ^{2x}\sin(3x)dx=\]
\[1/2(e ^{2x}\sin(3x))-3/4(e ^{2x}\cos(3x))-9/4\int\limits\limits_{}e ^{2x}\sin(3x)dx\]
Collect like terms:
\[\int\limits_{}e ^{2x}\sin(3x)dx+9/4\int\limits_{}e ^{2x}\sin(3x)dx=\]
\[1/2(e ^{2x}\sin(3x))-3/4(e ^{2x}\cos(3x))\]
so...
\[13/4\int\limits_{}e ^{2x}\sin(3x)dx=1/2(e ^{2x}\sin(3x))-3/4(e ^{2x}\cos(3x))\]
\[\int\limits\limits_{}e ^{2x}\sin(3x)dx=4/13[1/2(e ^{2x}\sin(3x))-3/4(e ^{2x}\cos(3x))]\]
\[\int\limits\limits_{}e ^{2x}\sin(3x)dx=2/13(e ^{2x}\sin(3x))-3/13e ^{2x}\cos(3x))+c\]

to check the answer just take the derivative of the right hand side and you will end up with the function in the intergrand on the left hand side, which is what you started with
Hope this helps