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Evaluate : `int_(0)^((pi)/(2)) sin^(2) x dx `.

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`int_(0)^((pi)/(2)) sin^(2) x *dx = int_(0) ^((pi)/(2)) ((1 - cos 2x)/(2)) dx `
` [ because cos 2x = 1 - 2 sin^(2) x] `
`= (1)/(2) int_(0)^((pi)/(2)) (1- cos 2x )dx`
` = (1)/(2) [ x - (sin 2x)/(2)]_(0) ^((pi)/(2))`
` = (1)/(2) [(pi)/(2) - (sin pi)/(2)]`
`therefore int_(0)^((pi)/(2)) sin^2 x. d x = 1/2xxpi/2=pi/4`

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