# Thread: [SOLVED] Evaluating Definite Integral (if exists)

1. ## [SOLVED] Evaluating Definite Integral (if exists)

Hi,
I'm I on the right track with this problem? Thanks in advance!

1. The problem statement, all variables and given/known data
Evaluate the definite Integral, if it exists

2. Relevant equations
integral (lower limit = 0, upper limit = square root of pi) of x cos (x^2) dx

3. The attempt at a solution
integral (lower limit = 0, upper limit = square root of pi) of x cos (x^2) dx

Let U = x^2

du/dx = 2x

du = 2xdx

dx = du / 2x

integral (lower limit = 0, upper limit = square root of pi) x cos u du/2x

(x/2 * - sin u 1/2x) + C

-1/4 [ sin u ] + c

-1/4 [ sin x^2 ]

-1/4 [ sin pi ] - [ 1/4 sin 0 ]

= 0

2. Originally Posted by sparky
Hi,
I'm I on the right track with this problem? Thanks in advance!

1. The problem statement, all variables and given/known data
Evaluate the definite Integral, if it exists

2. Relevant equations
integral (lower limit = 0, upper limit = square root of pi) of x cos (x^2) dx

3. The attempt at a solution
integral (lower limit = 0, upper limit = square root of pi) of x cos (x^2) dx

Let U = x^2

du/dx = 2x

du = 2xdx

dx = du / 2x

integral (lower limit = 0, upper limit = square root of pi) x cos u du/2x = (1/2) cosu du.
Integrating this gives (1/2) sin(u). Nou substitute u=x^2 and solve (1/2)sin(x^2) for the upper and lower limits.

(x/2 * - sin u 1/2x) + C

-1/4 [ sin u ] + c

-1/4 [ sin x^2 ]

-1/4 [ sin pi ] - [ 1/4 sin 0 ]

= 0
There is a small error. Look at the edited part and solve.

3. $\int_{0}^{\sqrt{\pi }} x*\cos (x^{2})(2xdx)=\frac{(\sin (x^2))}{2}=0-0=0$

You don't need a C with a definite integral and I am not sure where you obtained the 1/4

4. Originally Posted by dwsmith
$\int_{0}^{\sqrt{\pi }} x*\cos (x^{2})(2xdx)=\frac{(\sin (x^2))}{2}=0-0=0$

You don't need a C with a definite integral and I am not sure where you obtained the 1/4
Ok, thanks very much harish21 and dwsmith!