Integrating x^2sinx: Finding Limits and Solution

[Mo]

Homework Statement

Find the integral of:

Homework Equations

x^2 . sinx dx (With upper and lower limits of pi and 0 respectively.)

The Attempt at a Solution

Integrating for the first time:
-x^2 . cos x + 2 [integral of] x . cosx dx

After integrating a part of my first integration:
-x^2 . cosx + 2xsinx + 2cosx

----------------------------

I used the method of integration by parts where u = x^2 and dv = sinx.

I don't really need the final answer, i just want to see where i am going wrong with the actual integration.

Anything with trig in it seems to completely throw me

Thanks.

----------------------------

 

[marlon]

Homework Statement

Find the integral of:

Homework Equations

x^2 . sinx dx (With upper and lower limits of pi and 0 respectively.)

The Attempt at a Solution

Integrating for the first time:
-x^2 . cos x + 2 [integral of] x . cosx dx

After integrating a part of my first integration:
-x^2 . cosx + 2xsinx + 2cosx

----------------------------

I used the method of integration by parts where u = x^2 and dv = sinx.

I don't really need the final answer, i just want to see where i am going wrong with the actual integration.

Anything with trig in it seems to completely throw me

Thanks.

----------------------------

Mo, your solution (-x^2 . cosx + 2xsinx + 2cosx) is CORRECT :: approve

marlon

 

[Mo]

Thanks for the help. The thing is, if i input the limits of pi and 0, i get a completely different answer to that in the book!

The book gives the answer as ((pi^2) - 4)

Either i am substituting the limits incorrectly (likely) or the book's answer is incorrect (unlikely..)

Am I right in saying that first we substitute pi for x, get an answer, then substitute 0 for x and again get an answer.

We then subtract the our the latter answer from the first.

ie:

(-pi^2 . cos(pi) + 2(pi)sin(pi) + 2cos(pi)) - (2cos(0))

which will give us the final answer.

Regards.

ps: thanks for spotting that, cristo

 

[cristo]

ie:

(-pi^2 . cos(pi) + 2(pi)sin(pi) + 2cos(pi)) - (2cos(pi))

Probably a typo, but this should read

(-pi^2 . cos(pi) + 2(pi)sin(pi) + 2cos(pi)) - (2cos(0))

 

[marlon]

(-pi^2 . cos(pi) + 2(pi)sin(pi) + 2cos(pi)) - (2cos(pi))

which will give us the final answer.

Regards.

again correct...It must be a typo

cos(pi) = -1
sin(pi) = 0

i get

(-pi^2* (-1) + 2(pi)* 0 + 2* (-1)) - (2)
marlon

 

[Mo]

In that case the book was correct.

I was doing the substitutions using my calculator which was not in radians mode.

Thanks for your help though, much appreciated.