Posted by **CMM** on Thursday, January 27, 2011 at 7:37pm.

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?

- Calc. -
**helper**, Thursday, January 27, 2011 at 7:57pm
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))?

- Calc. -
**CMM**, Thursday, January 27, 2011 at 8:01pm
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.

- Calc. -
**helper**, Thursday, January 27, 2011 at 8:52pm
| x cos(x^2) dx

| = integral symbol

By half-angle 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!

- Calc. -
**helper**, Thursday, January 27, 2011 at 8:55pm
| x cos(x^2) dx

The first line should be

| x cos^2 x

I forgot to change it.

- Calc. -
**CMM**, Thursday, January 27, 2011 at 9:05pm
But the x is the only thing being squared I thought because it is within parentheses and then multiplied by xcos dx?

- Calc. -
**helper**, Thursday, January 27, 2011 at 9:09pm
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.

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