Calc.
posted by CMM on .
Evaluate the definite integral.
S b= sqrt(Pi) a= 0 xcos(x^2)dx
I'm not sure if this is right?
u= x^2
du= 2xdx
du/2= xdx
S (1/2)cos(u)
S (1/2)*sin(x^2)
[0.5 * sin(sqrt(Pi))^2]  [0.5 * sin(0)^2]
0  0 = 0, so zero is the answer?

S b= sqrt(Pi) a= 0 xcos(x^2)dx
What is "S b = sqrt(Pi) a= 0"
What are you integrating?
Is it this?
x cos(x^2) dx ?
from a = 0 to b = (sqrt(pi))? 
S is supposed to be the integral symbol and b is the upper bound and a is the lower bound. Yes, xcos(x^2) dx is it.

 x cos(x^2) dx
 = integral symbol
By halfangle cos^2 x = 1/2 (1 + cos 2x)
 x 1/2 (1 + cos 2x) dx
1/2  x + x cos 2x dx
1/2  x dx + 1/2  x cos 2x dx
1/4 x^2 + 1/2  x cos 2x dx
Integrate  x cos 2x dx by Integration by Parts
u = x, dv = cos 2x dx
du = dx, v = 1/2 sin 2x
1/4 x^2 + 1/2 (1/2 x sin 2x   1/2 sin 2x dx )
1/4 x^2 + 1/4x sin 2x  1/4 sin 2x dx)
u = 2x
du = 2 dx
1/2 du = dx
1/4 x^2 + 1/4x sin 2x  1/4 1/2 sin u du
1/4 x^2 + 1/4x sin 2x  1/8  sin u du
1/4 x^2 + 1/4x sin 2x  1/8 (cos u)+ C
1/4 x^2 + 1/4x sin 2x + 1/8 cos u + C
1/4 x^2 + 1/4x sin 2x + 1/8 cos 2x + C
You can't do it like you did above because
'cos^2 x' does not mean just the 'x' is squared.
cos^2 x means (cos x)^2
I did not do the a to b part. check back if you need help with that part.
Good Luck! 
 x cos(x^2) dx
The first line should be
 x cos^2 x
I forgot to change it. 
But the x is the only thing being squared I thought because it is within parentheses and then multiplied by xcos dx?

After all that, I confused myself and did the wrong problem I think.
IF your problem is
x cos(x^2) dx
just like your 2nd reply says, ignore the above.
I wrote down  x cos (x^2) but integrated
 x cos^2 x dx
So, your very first post IS correct if it is cos x^2 and not cos^2 x.
BIG difference in the integration!!
Well, now you know how to integrate x cos^2 x if you have to in the future.