Register to reply 
Evaluating the integral, correct? 
Share this thread: 
#1
Jan1808, 09:36 PM

P: 62

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 


#2
Jan1808, 09:45 PM

P: 1,756

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 


#3
Jan1808, 10:00 PM

HW Helper
P: 1,664

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... 


#4
Jan1808, 10:08 PM

P: 62

Evaluating the integral, correct?
i heard that i should do parts with xsinmx / m, should i do that?



#5
Jan1808, 10:13 PM

HW Helper
P: 1,664




#6
Jan1808, 10:15 PM

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?



#7
Jan1808, 10:23 PM

HW Helper
P: 1,664

(1/m) · integral[ x sin(mx) ] dx . Now u = x and dv = sin(mx) dx . You can append the multiplicative constant (1/m) afterwards... 


#8
Jan1808, 10:37 PM

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 


#9
Jan1808, 10:45 PM

P: 1,756

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!!! 


#10
Jan1808, 11:01 PM

P: 1,756

Ok, I'm officially done typing! Sorry about that, had too many typographical errors.



#11
Jan1808, 11:02 PM

P: 62

[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 


#13
Jan1808, 11:07 PM

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 :(



#15
Jan1808, 11:19 PM

P: 62

[integral] e^x cos 2x dx
u = e^x du = e^x dv = cos 2x v = sin 2x / 2 


#16
Jan1808, 11:22 PM

P: 1,756

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 ... [tex]\int e^{x}\cos{2x}dx[/tex] Right? 


#17
Jan1808, 11:28 PM

P: 1,756

[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! 


#18
Jan1808, 11:34 PM

P: 62

ok so [tex]\frac{1}{2}\int e^{x}\sin{2x}dx[/tex] turns into 1/2e^xcos2x + c?



Register to reply 
Related Discussions  
Help Evaluating an Integral  Calculus & Beyond Homework  2  
Evaluating an integral  Calculus & Beyond Homework  1  
Evaluating an integral  Calculus & Beyond Homework  5  
Evaluating a cot integral  Precalculus Mathematics Homework  3  
Evaluating Integral  Calculus & Beyond Homework  4 