Question

I don't believe all C's are created equal
\[\int x^{2}cosmxdx\]
u= x^2 dv=cosmx dx
du=2x dx v=(1/m)sinmx
\[\frac{x^{2}}{m}sin(mx)-\frac{2}{m}\int xsin(mx) dx+C\]
u=x dv=sinmxdx
du=dx v=-(1/m)cosmx
\[-\frac{x}{m}cos(mx) + \frac{1}{m}\int cos(mx) dx \]
\[(-\frac{x}{m}cos(mx) +\frac{1}{m^{2}}sin(mx) +C)\]
\[\frac{x^{2}}{m}sin(mx)-\frac{2}{m}(-\frac{x}{m}cos(mx) +\frac{1}{m^{2}}sin(mx) +C)+C\]
\[\frac{x^{2}}{m}sin(mx)+\frac{2x}{m^{2}}cos(mx) -\frac{2}{m^{3}}sin(mx) -\frac{2}{m}C+C\]

Answer 1

how are my C's looking?

Answer 2

why do you not use tabulars method

Answer 3

x^2 goes to 0

Answer 4

i was just thinking about that ...:)

Answer 5

I have never used Tabulars method before, but I'm willing to learn

Answer 6

i do believe there is a mistake as you shouldn't get that large of an integration

Answer 7

it's a type of by parts method where your u goes to 0

Answer 8

after so many derivatives

Answer 9

so you make u = x^2 and dv = cos(mx)

Answer 10

then you take as many derivatives as it takes to make ti got o0