求助一下这道不定积分题怎么做

更新时间：2021-12-06 09:33:45 117次访问

∫[x²/√(a²-x²)]dx

=-0.5∫[x/√(a²-x²)]d(a²-x²)

=-∫xd√(a²-x²)

=-x√(a²-x²)+∫√(a²-x²)dx

=-x√(a²-x²)+∫[(a²-x²)/√(a²-x²)]dx

=-x√(a²-x²)+a²∫dx/√(a²-x²)-∫[x²/√(a²-x²)]dx

=0.5[-x√(a²-x²)+a²∫d(x/a)/√(1-x²/a²)]

=-0.5x√(a²-x²)+0.5a²arcsin(x/a)+c