求不定积分∫x^2⼀√(a^2-x^2)dx=?

2024-11-15 10:32:11
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回答1:

令x=asint,则dx=acost dt ∫x²/√如陵(a²-x²) dx=∫a²sin²t/(acost)·acostdt=a²∫sin²t dt=a²∫(1-cos2t)/2 dt=a²∫1/2dt-a²∫cos2tdt=a²渣纯戚t/2-1/2·a²裤皮sin2t+C=1/2·a²arcsin(x/a)-x·√(a²-x²)+C