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प्रश्न
Evaluate: `int_(pi/6)^(pi/3) sin^2 x "d"x`
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उत्तर
`int_(pi/6)^(pi/3) sin^2 x "d"x = int_(pi/6)^(pi/3) ((1 - cos 2x)/2) "d"x`
= `1/2[int_(pi/6)^(pi/3) "d"x - int_(pi/6)^(pi/3) cos 2x "d"x]`
= `1/2[[x]_(pi/3)^(pi/6) - [(sin 2x)/2]_(pi/6)^(pi/3)]`
= `1/2[(pi/3 - pi/6) - 1/2(sin (2pi)/3 - sin pi/3)]`
= `1/2[pi/6 - 1/2(sqrt(3)/2 - sqrt(3)/2)]`
= `1/2[pi/6 - 1/2 (0)]`
= `pi/12`
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