Question

(sinx)^7的积分?

Answer 1

原式=-∫（sinx)^6dcosx
=-∫[1-(cosx)^2]^3d(cosx)
设t=cosx,
原式=-∫(1-t^2)^3dt
=-∫(1-3t^2+3t^4-t^6)dt
=-(t-t^3+3t^5/5-t^7/7)+C
=-[cosx-(cosx)^3+3(cosx)^5/5-(cosx)^7/7]+C.