## Denebel 3 years ago evaluate the integral analytically.

1. Denebel

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2. shinigami1m

use integration by parts

3. Zarkon

the integrand is positive on 0 to pi/2 so the answer has to be a positive number

4. Denebel

The answer is (pi^2/8) - (1/2)

5. Denebel

according to the back of the book

6. Zarkon

7. Denebel

Ok, how do I get there?

8. Zarkon

integration by parts ;)

9. ash2326

sorry it'd be -x^2 cos x + $$\int\limits_{0}^{\pi/2} 2x \cos xdx$$ -x^2 cos x+2x sin x-\(\int\ limits_{0}^{pi/2} 2 sin x dx so we get -x^2 cos x+ 2x sin x+2 cos x 2x sin x- cos x(x^2 -2) now insert the limits [ 2* pi/2 * 1-0]-[0-1*(0-2)] pi -2

10. Zarkon

that is correct if it was $x^2\sin(x)$ but it is $x^2\sin(2x)$

11. ash2326

sorry i thought it's x^2 sin x let me put the correct one -x^2 cos 2x /2+ ∫ 0 π/2 xcos2xdx -x^2 cos 2x /2 + x sin 2x/2 -∫ 0 π/2 sin2x/2dx -x^2 cos 2x/2 + x sin 2x /2 +cos 2x /4 now inserting the limits [pi^2/8+0-1/4]-[1/4] pi^2/8-1/2 π^2 /8-1/2

12. Denebel

Thanks!!