Question

高数不定积分题目求解答

Answer 1

原式 = -∫(sinx)^2(cosx)^2d(cosx)
=-∫[1-(cosx)^2](cosx)^2d(cosx)
=-∫[(cosx)^2-(cosx)^4]d(cosx)
=-∫(t^2-t^4)dt
=t^5/5-t^3/3+C
=(cosx)^5/5-(cosx)^3/3+C

Answer 2

(sinx)^3 (cosx)^2 dx = -(sinx)^2 (cosx)^2dcosx =[(cosx)^2-1] (cosx)^2 ]dcosx
= [(cosx)^4-(cosx)^2 ]dcosx
=d [(cosx)^5/5-(cosx)^3/3 ]
所以原不定积分＝1/5*(cosx)^5-1/3*(cosx)^3 + C (C是常数）