> 数学 >
用第二换元法求不定积分:∫x^2dx/√1-x^2
人气:273 ℃ 时间:2020-03-15 15:46:58
解答
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
推荐
猜你喜欢
© 2024 79432.Com All Rights Reserved.
电脑版|手机版