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Question
Evaluate the following:
`int_0^(pi/2) sin^2x cos^4 x "d"x`
Solution
`int_0^(pi/2) sin^2x cos^4 x "d"x = int_0^(pi/2) (1 - cos^2x) cos^4x "d"x`
= `int_0^(pi/2) (sin^4x - cos^6x) "d"x`
= `int_0^(pi/2) cos^4x "d"x - int_0^(pi/2) cos^6x "d"x`
= `(3/4xx 1/2 xx pi/2) - (5/6 xx 3/4 xx 1/2 xx pi/2)`
= `(3pi)/16 - (5pi)/32`
= `pi/16[3 - 5/2]`
= `pi/16[(6 - 5)/2]`
= `pi/16(1/2)`
= `pi/32`
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