Question

求∫x根号（1-x^2）dx的不定积分

Answer 1

x=sina
dx=cosada
√(1-x²)=cosa
原式=∫sinacosa*cosada
=∫sinacos²ada
=-∫cos²adcosa
=-cos³a/3+C
=-[√(1-x²)]²/3+C
=-(1-x²)√(1-x²)/3+C