Question

求定积分 上限为1/2，下限0，∫arccosxdx

求详细过程，谢谢啦

Answer 1

∫(0->1/2) arccosx dx
= x * arccosx - ∫(0->1/2) x d(arccosx)
= (1/2) * arccos(1/2) - ∫(0->1/2) x * -1/√(1-x²) dx
= π/6 + (1/2)∫(0->1/2) 1/√(1-x²) d(x²)
= π/6 - (1/2)∫(0->1/2) 1/√(1-x²) d(1-x²)
= π/6 - (1/2) * 2√(1-x²)
= π/6 - [√(1-(1/2)²) - 1]
= 1 - √3/2 + π/6