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Question
Evaluate : `int _0^(pi/2) "sin"^ 2 "x" "dx"`
Solution
`int _0^(pi/2) "sin"^ 2 "x" "dx"`
i = `int _0^(pi/2) (1 - cos 2"x" "dx") /2` [∵ 1 - 2 cos2 θ = 2 sin 2 θ]
`["x"/2 - ("sin"2"x")/4]_0^(pi/2)`
=`(pi/4 -("sin" pi )/4) - (0 - 0)`
=`pi/4`
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