# Evaluating the integral, correct?

by Zack88
Tags: correct, evaluating, integral
 P: 62 1. The problem statement, all variables and given/known data Evaluate the integral $$\int x^2 \cos mx dx$$ 2. Relevant equations Evaluating the integral, correct? 3. The attempt at a solution $$u = x^2$$ $$du = 2x$$ $$dv = \cos mx$$ $$v= \frac {\sin mx }{m}$$ (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
 P: 1,754 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 :-] http://physicsforums.com/showthread.php?t=8997
 HW Helper P: 1,662 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...
 P: 62 Evaluating the integral, correct? i heard that i should do parts with xsinmx / m, should i do that?
HW Helper
P: 1,662
 Quote by Zack88 i heard that i should do parts with xsinmx / m, should i do that?
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.
 P: 62 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?
HW Helper
P: 1,662
 Quote by Zack88 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) · integral[ x sin(mx) ] dx . Now u = x and dv = sin(mx) dx . You can append the multiplicative constant (1/m) afterwards...
 P: 62 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
 P: 1,754 Ok let's start from scratch. $$I=\int x^2\cos{mx}dx$$ $$u=x^2$$ $$du=2xdx$$ $$dV=\cos{mx}dx$$ $$V=\frac{1}{m}\sin{mx}$$ $$I=\frac{x^2}{m}\sin{mx}-\frac{2}{m}\int x\sin{mx}dx$$ Now we have to do Parts again. $$u=x$$ $$du=dx$$ $$dV=\sin{mx}dx$$ $$V=\frac{-1}{m}\cos{mx}$$ $$I=\frac{x^2}{m}\sin{mx}-\frac{2}{m}\left(\frac{-x}{m}\cos{mx}+\frac{1}{m}\int\cos{mx}dx\right)$$ Now you can easily evaluate this Integral!!!
 P: 1,754 Ok, I'm officially done typing! Sorry about that, had too many typographical errors.
 P: 62 $$I=\frac{x^2}{m}\sin{mx}-\frac{2}{m}\left(\frac{-x}{m}\cos{mx}+\int\cos{mx}dx\right)$$ ok where did the last $$\int\cos{mx}dx\right)$$ come from
P: 1,754
 Quote by Zack88 $$I=\frac{x^2}{m}\sin{mx}-\frac{2}{m}\left(\frac{-x}{m}\cos{mx}+\int\cos{mx}dx\right)$$ ok where did the last $$\int\cos{mx}dx\right)$$ come from
By doing Parts again to evaluate ...

$$\int x\sin{mx}dx$$
 P: 62 ok i see now, thank you, sad part is im not done with this question and i have one more just like it :(
P: 1,754
 Quote by Zack88 ok i see now, thank you, sad part is im not done with this question and i have one more just like it :(
Post it and we'll work on it step by step.
 P: 62 [integral] e^-x cos 2x dx u = e^-x du = -e^-x dv = cos 2x v = sin 2x / 2
 P: 1,754 Can you check post #9 again http://physicsforums.com/showpost.ph...42&postcount=9 I forgot to type in the constant. Anyways, your new problem is ... $$\int e^{-x}\cos{2x}dx$$ Right?
 P: 1,754 $$I=\int e^{-x}\cos{2x}dx$$ $$u=e^{-x}$$ $$du=-e^{-x}dx$$ $$dV=\cos{2x}dx$$ $$V=\frac{1}{2}\sin{2x}$$ $$I=\frac{1}{2}e^{-x}\sin{2x}+\frac{1}{2}\int e^{-x}\sin{2x}dx$$ 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!
 P: 62 ok so $$\frac{1}{2}\int e^{-x}\sin{2x}dx$$ turns into -1/2e^-xcos2x + c?

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