问题描述：
设f'(x)=arcsinx^2,且f(1)=0,求I=S(0,1)f(x)dx
S是不定积分号,０是积分下限,１是积分上线
问题解答：
我来补答
∫(0→1) f(x) dx
= xf(x) |(0→1) - ∫(0→1) xf'(x) dx
= f(1) - ∫(0→1) x(arcsinx)² dx
= - ∫(0→1) x(arcsinx)² dx
= (- 1/2)∫(0→1) (arcsinx)² d(x²)
= (- 1/2)(xarcsinx)² |(0→1) + (1/2)∫(0→1) x² d(arcsinx)²
= (- 1/2)(π/2)² + ∫(0→1) x²arcsinx/√(1 - x²) dx
= - π²/8 + ∫(0→1) x²arcsinx/√(1 - x²) dx
令x = siny ==> dx = cosy dy
= - π²/8 + ∫(0→π/2) ysin²y/cosy • cosy dy
= - π²/8 + ∫(0→π/2) y • (1 - cos2y)/2 dy
= - π²/8 + ∫(0→π/2) y/2 dy - (1/2)∫(0→π/2) ycos2y dy
= - π²/8 + y²/4 |(0→π/2) - (1/2)(1/2)∫(0→π/2) y d(sin2y)
= - π²/8 + (1/4)(π/2)² - (1/4)ysin2y |(0→π/2) + (1/4)∫(0→π/2) sin2y dy
= - π²/8 + π²/16 + (1/4)(- 1/2)cos2y |(0→π/2)
= - π²/16 - (1/8)[(- 1) - 1]
= (4 - π²)/16
