請問各位數(shù)學(xué)天才關(guān)于∫(arcsinx)²dx=?的這兩種做法都正確嗎
令t=arcsinx,則x=sint,dx=cost dt
∫(arcsinx)²dx
=∫ t²·cost dt
=t²·sint-∫ 2t·sint dt
=t²·sint+∫ 2t·d(cost)
=t²·sint+2tcost-∫ 2cost dt
=t²·sint+2tcost-2sint+C
=x·arcsin²x+2arcsinx·√(1-x²)-2x+C
∫(arcsinx)²dx
=x(arcsinx)²-∫2xarcsinx/√(1-x^2)dx
=x(arcsinx)²+∫(arcsinx)/√(1-x^2)d(1-x^2)
=x(arcsinx)²+(2/3)∫(arcsinx) d (1-x^2)^(3/2)
=x(arcsinx)²+(2/3)(arcsinx)(1-x^2)^(3/2)-(2/3)∫(1-x^2)dx
=x(arcsinx)²+(2/3)(arcsinx)(1-x^2)^(3/2)-(2/3)x+(2/9)x^3+C
=x(arcsinx)²+(2/3)(arcsinx)(1-x^2)^(3/2)-(2/3)x+(2/9)x^3+C
兩次應(yīng)用分部積分法,就可以解出來了