Evaluating the integral, correct?

[Zack88]

Homework Statement

Evaluate the integral

\int x^2 \cos mx dx

Homework Equations

Evaluating the integral, correct?

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

 

[rocomath]

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

 

[dynamicsolo]

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...

 

[Zack88]

i heard that i should do parts with xsinmx / m, should i do that?

 

[dynamicsolo]

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.

 

[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?

 

[dynamicsolo]

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...

 

[Zack88]

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 I am lost

 

[rocomath]

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!

 

[rocomath]

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

 

[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

 

[rocomath]

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

 

[Zack88]

ok i see now, thank you, sad part is I am not done with this question and i have one more just like it :(

 

[rocomath]

ok i see now, thank you, sad part is I am not done with this question and i have one more just like it :(

Post it and we'll work on it step by step.

 

[Zack88]

[integral] e^-x cos 2x dx

u = e^-x
du = -e^-x

dv = cos 2x
v = sin 2x / 2

 

[rocomath]

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 ...

\int e^{-x}\cos{2x}dx

Right?

 

[rocomath]

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!

 

[Zack88]

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

 

[rocomath]

You're supposed to have "2" Parts just like any Integration by Parts.

What is the other term?

 

[Zack88]

so i don't take the integral just yet? i do another integration by parts?

 

[rocomath]

so i don't take the integral just yet? i do another integration by parts?

Evaluate \frac{1}{2}\int e^{-x}\sin{2x}dx how you normally would.

 

[Zack88]

u = e^-x
du = -e^-x

dv = sin 2x
v= -1/2 cos 2x

 

[rocomath]

Yes, so now you've solved it?

 

[Zack88]

not quite, I am not sure how to put it all together I know i'll have 1/2e^-xsin2x but I am not sure how to put in the new u, du, dv, and v

 

[rocomath]

not quite, I am not sure how to put it all together I know i'll have 1/2e^-xsin2x but I am not sure how to put in the new u, du, dv, and v

Look at what I did in post 9.

 

[Zack88]

ok I am not done yet but have so far...

1/2e^-xsin2x + 1/2 (1/2 e^-x cos2 x

and that is where i stopped i didnt understand how you got + 1/m [integra] cosmx for post 9

 

[rocomath]

By Parts.

uV-\int Vdu

 

[Zack88]

lol sorry i saw (\frac{-x}{m}\cos{mx}+\frac{1}{m}\int\cos{mx}dx\right) as two different things

 

[Zack88]

so now i have 1/2 e^-x sin 2x + 1/2 (1/2 e^-x cos 2x + [integral] 1/2 e^-x cos 2x)

 

[Zack88]

so now do I finally take the integral (no more parts) to come out with the final answer?