如题所述
第1个回答 2023-09-29
求不定积分∫arcsinxarccosxdx
解:令arcsinx=u,则x=sinu;dx=cosudu;arccosx=π/2-arcsinx=π/2-u;代入原式得:
原式=∫[u(π/2-u)cosudu=(π/2)∫ucosudu-∫u²cosudu=(π/2)∫ud(sinu)-∫u²dsinu
=(π/2)[usinu-∫sinudu]-[u²sinu-2∫usinudu]=(π/2)(usinu+cosu)-u²sinu-2∫ud(cosu)
=(π/2)(usinu+cosu)-u²sinu-2(ucosu-∫cosudu)
=(π/2)(usinu+cosu)-u²sinu-2(ucosu-sinu)+C
=[2+(π/2)u-u²]sinu+[(π/2)-2u]cosu+C
=[2+(π/2)arcsinx-(arcsinx)²]x-[(π/2)-2arcsinx]cos(arcsinx)+C
=[2+(π/2)arcsinx-(arcsinx)²]x-[(π/2)-2arcsinx]√(1-x²)+C
解:令arcsinx=u,则x=sinu;dx=cosudu;arccosx=π/2-arcsinx=π/2-u;代入原式得:
原式=∫[u(π/2-u)cosudu=(π/2)∫ucosudu-∫u²cosudu=(π/2)∫ud(sinu)-∫u²dsinu
=(π/2)[usinu-∫sinudu]-[u²sinu-2∫usinudu]=(π/2)(usinu+cosu)-u²sinu-2∫ud(cosu)
=(π/2)(usinu+cosu)-u²sinu-2(ucosu-∫cosudu)
=(π/2)(usinu+cosu)-u²sinu-2(ucosu-sinu)+C
=[2+(π/2)u-u²]sinu+[(π/2)-2u]cosu+C
=[2+(π/2)arcsinx-(arcsinx)²]x-[(π/2)-2arcsinx]cos(arcsinx)+C
=[2+(π/2)arcsinx-(arcsinx)²]x-[(π/2)-2arcsinx]√(1-x²)+C