1. The problem statement, all variables and given/known data Evaluate the integral [tex]\int x^2 \cos mx dx[/tex] 2. Relevant equations Evaluating the integral, correct? 3. The attempt at a solution [tex]u = x^2[/tex] [tex]du = 2x[/tex] [tex]dv = \cos mx[/tex] [tex]v= \frac {\sin mx }{m}[/tex] (x^2)(sin mx / m) - [integral] (sin mx / m)(2x) (x^2)(sin mx / m) - 2 [integral] (sin mx / m) (x) [My final answer:] (x^2)(sin mx / m) + 2 (cos mx / m) + c
Hurts the eyes. If you plan on coming here for help on a regular basis, hopefully you will take the time to learn how to type in LaTeX :-] https://www.physicsforums.com/showthread.php?t=8997
I believe you lost a term in evaluating the second term integral in this: (x^2)(sin mx / m) - 2 [integral] (sin mx / m) (x) . If you differentiate your final result, (x^2)(sin mx / m) + 2 (cos mx / m) + c , you don't cancel out the additional terms beyond the original integrand...
Yes, you'll need to do that. In general, integration of functions defined by polynomials times sin kx, cos kx, or e^kx need multiple stages of integration by parts.
ok with doing parts again confuses me since i have xsinmx /m. The u should it be m? but then what is the derivative of m?
The 'm' is a constant, so that will not be involved in the integration. Make the integral (1/m) ยท integral[ x sin(mx) ] dx . Now u = x and dv = sin(mx) dx . You can append the multiplicative constant (1/m) afterwards...
ok so now I have x^2/m sinmx + 2/m [integral] xsinmx u = x du = dx dv = sin mx v = 1/m cos mx and now im lost
Ok let's start from scratch. [tex]I=\int x^2\cos{mx}dx[/tex] [tex]u=x^2[/tex] [tex]du=2xdx[/tex] [tex]dV=\cos{mx}dx[/tex] [tex]V=\frac{1}{m}\sin{mx}[/tex] [tex]I=\frac{x^2}{m}\sin{mx}-\frac{2}{m}\int x\sin{mx}dx[/tex] Now we have to do Parts again. [tex]u=x[/tex] [tex]du=dx[/tex] [tex]dV=\sin{mx}dx[/tex] [tex]V=\frac{-1}{m}\cos{mx}[/tex] [tex]I=\frac{x^2}{m}\sin{mx}-\frac{2}{m}\left(\frac{-x}{m}\cos{mx}+\frac{1}{m}\int\cos{mx}dx\right)[/tex] Now you can easily evaluate this Integral!!!
[tex]I=\frac{x^2}{m}\sin{mx}-\frac{2}{m}\left(\frac{-x}{m}\cos{mx}+\int\cos{mx}dx\right)[/tex] ok where did the last [tex]\int\cos{mx}dx\right)[/tex] come from
ok i see now, thank you, sad part is im not done with this question and i have one more just like it :(
Can you check post #9 again https://www.physicsforums.com/showpost.php?p=1576142&postcount=9 I forgot to type in the constant. Anyways, your new problem is ... [tex]\int e^{-x}\cos{2x}dx[/tex] Right?
[tex]I=\int e^{-x}\cos{2x}dx[/tex] [tex]u=e^{-x}[/tex] [tex]du=-e^{-x}dx[/tex] [tex]dV=\cos{2x}dx[/tex] [tex]V=\frac{1}{2}\sin{2x}[/tex] [tex]I=\frac{1}{2}e^{-x}\sin{2x}+\frac{1}{2}\int e^{-x}\sin{2x}dx[/tex] Ok done typing. Evaluate your current Integral, notice that your Original Integral reappears? Just move it to the left then divide by the constant and you're done!