用第二换元法求不定积分:∫x^2dx/√1-x^2
问题描述：
用第二换元法求不定积分:∫x^2dx/√1-x^2
答
令：x=sint
∫x^2dx/√1-x^2
=∫sin^2t costdt /cost
=∫sin^2t dt
=1/2∫(1-cos2t)dt
=t/2-sin2t/4 +c
=t/2-sintcost/2+c
=1/2[arcsinx - x√1-x^2]+c