Help with integration problems?

Question

int e^2x*cos(x/3)dx, int x^2lnx dx limit from 1 to e

Answer 1

∫e^(2x) cos (x/3) dx

For this problem, we will use integration by parts twice to obtain an expression for this integral in terms of the same integral, and then simple algebra will suffice to provide us with the general integral. First let u=e^(2x), du = 2e^(2x) dx, v=3 sin (x/3), dv = cos (x/3) dx. So:

∫e^(2x) cos (x/3) dx = 3 e^(2x) sin (x/3) - ∫6 e^(2x) sin (x/3) dx + C

Now let u=6 e^(2x), du=12e^(2x) dx, v= -3 cos (x/3), dv = sin (x/3) dx. Integrating by parts again:

∫e^(2x) cos (x/3) dx = 3 e^(2x) sin (x/3) + 18 e^(2x) cos (x/3) - 36∫e^(2x) cos (x/3) dx + C

Now adding 36 ∫e^(2x) cos (x/3) dx to both sides:

37∫e^(2x) cos (x/3) dx = 3 e^(2x) sin (x/3) + 18 e^(2x) cos (x/3) + C

Dividing by 37:

∫e^(2x) cos (x/3) dx = (3 e^(2x) sin (x/3) + 18 e^(2x) cos (x/3))/37 + C

And we are done with the first problem:

For the second, it can be done through a much more straightforward integration by parts. We start with:

[1, e]∫x² ln x dx

Let u=ln x, du=1/x dx, v=x³/3, dv=x² dx. Then we have:

x³ ln x/3 |[1, e] - [1,e]∫x²/3 dx

Evaluating:

e³ ln e/3 - 1 ln 1/3 - [1,e]∫x²/3 dx
e³/3 - [1, e]∫x²/3 dx

Integrating:

e³/3 - x³/9 |[1, e]

Evaluating:

e³/3 - e³/9 + 1/9
(2e³+1)/9

And we are done.

Answer 2

indispensable with the aid of areas is: ?udv=uv-?vdu so after substitution and putting u=x and dv=F'(x), ?xF'(x)dx=xF(x)-?F(x) and because ?F(x)=G(x) with the aid of fact F(x)=G'(x) ?xF'(x)dx=xF(x)-G(x).