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shinigami1m
 3 years ago
Best ResponseYou've already chosen the best response.0use integration by parts

Zarkon
 3 years ago
Best ResponseYou've already chosen the best response.0the integrand is positive on 0 to pi/2 so the answer has to be a positive number

Denebel
 3 years ago
Best ResponseYou've already chosen the best response.0The answer is (pi^2/8)  (1/2)

Denebel
 3 years ago
Best ResponseYou've already chosen the best response.0according to the back of the book

Denebel
 3 years ago
Best ResponseYou've already chosen the best response.0Ok, how do I get there?

ash2326
 3 years ago
Best ResponseYou've already chosen the best response.2sorry 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 * 10][01*(02)] pi 2

Zarkon
 3 years ago
Best ResponseYou've already chosen the best response.0that is correct if it was \[x^2\sin(x)\] but it is \[x^2\sin(2x)\]

ash2326
 3 years ago
Best ResponseYou've already chosen the best response.2sorry 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+01/4][1/4] pi^2/81/2 π^2 /81/2
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