∫x^2/(1-x^2)^1/2 dx 用第二类换元法求不定积分过程,
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![∫x^2/(1-x^2)^1/2 dx 用第二类换元法求不定积分过程,](/uploads/image/z/2694210-42-0.jpg?t=%E2%88%ABx%5E2%2F%281-x%5E2%29%5E1%2F2+dx+%E7%94%A8%E7%AC%AC%E4%BA%8C%E7%B1%BB%E6%8D%A2%E5%85%83%E6%B3%95%E6%B1%82%E4%B8%8D%E5%AE%9A%E7%A7%AF%E5%88%86%E8%BF%87%E7%A8%8B%2C)
∫x^2/(1-x^2)^1/2 dx 用第二类换元法求不定积分过程,
∫x^2/(1-x^2)^1/2 dx 用第二类换元法求不定积分过程,
∫x^2/(1-x^2)^1/2 dx 用第二类换元法求不定积分过程,
这个不用二类换元法反而简单些吧
设x=sint,dx=costdt,√(1-x^2)=cost
原式=∫sin²t/cost*costdt
=∫sin²tdt
=∫(1-cos2t)/2dt
=1/2∫dt-1/2∫cos2tdt
=t/2+1/4∫cos2td2t
=t/2+1/4*sin2t+C
=t/2+1/2*sintcost+C
=1/2*arcsinx+1/2*x/√(1-x^2)+C