Integrate sin(x)cos(x)^2 from 0 to π/2
Use substitution u=cos(x) resulting in du=-sin(x)dx: ∫0π/2sin(x)cos(x)^2dx = ∫0π/2-u^2du = [-1/3 u^3]x=0x=π/2 = [-1/3 cos(x)^3]0π/2 = (-1/3 cos(π/2)^3) - (-1/3 cos(0)^3) = (-1/3 0^3 ) - (-1/3 1^3) = 0 + 1/3 = 1/3
BS
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