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# Evaluate: ?x^2 ?sin?^2 x dx

Asked by Manoj 16th March 2013, 5:54 AM

write sin^2(x) = [1 - cos(2x)]/2

=1/2?x^2 (1- cos(2x)) dx

=1/2?x^2 dx - 1/2?x^2 cos(2x) dx

=(1/6)x^3 - ?(1/2)x^2 cos(2x) dx

integrate by parts twice

First function= (1/2)x^2 and second function = cos(2x)

?(1/2)x^2 cos(2x) dx = (1/4)x^2 sin(2x) - ?(1/2)x sin(2x) dx

again integrate by parts

first function = (1/2)x, second function = sin(2x)

?(1/2)x^2 cos(2x) dx = (1/4)x^2 sin(2x) + (1/4) x cos(2x) - 1/4? cos(2x) dx

=(1/4)x^2 sin(2x) + (1/4) x cos(2x) - (1/8)sin(2x)+c

so final answer is

(1/6)x^3 - (1/4)x^2 sin(2x) - (1/4) x cos(2x) + (1/8)sin(2x)+c
Answered by Expert 16th March 2013, 6:14 AM
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