1、求∫xcos(x/2)dx的不定积分2、用第二类换元法求∫x^2√（x-3)dx的积分!

Question

1、求∫xcos(x/2)dx的不定积分 2、用第二类换元法求∫x^2√（x-3)dx的积分!

Answer 1

∫ x*cos(x/2) dx = 2∫ x*cos(x/2) d(x/2) = 2∫ x dsin(x/2)
= 2xsin(x/2) - 2∫ sin(x/2) dx
= 2xsin(x/2) - 4∫ sin(x/2) d(x/2)
= 2xsin(x/2) + 4cos(x/2) + C
∫ √(x-3) dx,用第一类换元法较好
令u = √(x-3),du = dx/[2√(x-3)]
原式= 2∫ (u?+3)?du
= 2∫ (u^6+6u^4+9u?) du
= (2/7)u^7 + (12/5)u^5 + 6u?+ C
= (2/7)(x-3)^(7/2) + (12/5)(x-3)^(5/2) + 6(x-3)^(3/2) + C
= (2/35)(5x?+12x+24)(x-3)^(3/2) + C