先用分部积分法，然后再换元，另x＝sint，然后就能求出来了，如图
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第1个回答 2018-12-12
使用分部积分法
=0.5arcsinxd(x^2)
=0.5*x^2arcsinx-0.5x^2d(arcsinx)
答案：1/2*x^2*asin(x)+1/4*x*(1-x^2)^(1/2)-1/4*asin(x)
=0.5arcsinxd(x^2)
=0.5*x^2arcsinx-0.5x^2d(arcsinx)
答案：1/2*x^2*asin(x)+1/4*x*(1-x^2)^(1/2)-1/4*asin(x)
第2个回答 2018-12-12
答案在纸上
第3个回答 2018-12-12
解:原式=1/2∫arcsinxdx^2
=1/2(x^2arcsinx-∫x^2darcsinx)
=1/2(x^2arcsinx-∫(x^2/√1-x^2)dx)
设x=cost(0<t<π)
则∫(x^2/√ 1-x^2)dx
=∫[cos^2(t)/sint]dcost
=-∫cos^2(t)dt
=-1/2∫(1+cos2t)dt
=-1/2(∫dt+1/2∫cos2td2t)
=-1/2(t+1/2sin2t)+C
所以原式=1/2(x^2arcsinx)+1/4arccosx+
+1/4(x√1-x^2)+C
=1/2(x^2arcsinx-∫x^2darcsinx)
=1/2(x^2arcsinx-∫(x^2/√1-x^2)dx)
设x=cost(0<t<π)
则∫(x^2/√ 1-x^2)dx
=∫[cos^2(t)/sint]dcost
=-∫cos^2(t)dt
=-1/2∫(1+cos2t)dt
=-1/2(∫dt+1/2∫cos2td2t)
=-1/2(t+1/2sin2t)+C
所以原式=1/2(x^2arcsinx)+1/4arccosx+
+1/4(x√1-x^2)+C
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