Calculation of definite integral ∫ (x ^ (1 / 2) * sin2x) DX Previous question hint:u=x/2 - 1/8 du/dx=sin^2 (2x)

Calculation of definite integral ∫ (x ^ (1 / 2) * sin2x) DX Previous question hint:u=x/2 - 1/8 du/dx=sin^2 (2x)

This problem needs to be divided into two parts. It's troublesome. It's hard to understand it. If you don't understand it, please email me. First, I = ∫ (x ^ (1 / 2) * sin2x) DX part: - cos (2x) x ^ 1 / 2 / 2 + 1 / 2 ∫ cos (2x) DX / x ^ 1 / 2, right? Let V '= sin (x), u = x ^ 1 / 2, the second part I = - cos (2x) x ^ 1 / 2

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