求x·arccosx的不定积分!

题目

求x·arccosx的不定积分!

答案

法一：
先用分部积分
∫x·arccosx dx
=x²/2·arccosx-∫x²/2·[-1/√(1-x²)] dx
=x²/2·arccosx+1/2 ∫x²/√(1-x²) dx
下面求 ∫x²/√(1-x²) dx
令sint=x,则dx=cost dt
∫x²/√(1-x²) dx
=∫sin²t/cost ·costdt
=∫sin²t dt
=∫(1-cos2t)/2 dt
=t-1/4·sin2t+C
=arcsinx-1/2·x√(1-x²)+C
∴ ∫x·arccosx dx=x²/2·arccosx+1/2·arcsinx-1/4·x√(1-x²)+C
法二：
令arccosx=t,则x=cost,dx=-sint dt
∫x·arccosx dx
=∫cost·t·(-sint)dt
=-1/2∫sin2t·t dt
=-1/2[(-cos2t)/2·t+1/2∫cos2tdt]
=-1/2[(-cos2t)/2·t+1/4·sin2t]+C
=1/4·cos2t·t-1/8·sint2t+C
=x²/2·arccosx+1/2·arcsinx-1/4·x√(1-x²)+C