sin^5cos^2dx不定积分

Answer 1

原式=-积分(1-cos^2x)^2*cos^2x dcosx
令y=cosx
=-积分(1-y^2)^2 * y^2 dy
=-积分(1+y^4-2y^2)y^2 dy
=-积分(y^2+y^6-2y^4)dy
=-(y^3/3 + y^7/7 - 2y^5/5) +C
=-y^3(1/3+y^4/7-2y^2/5) +C
=-cos^3x(1/3+cos^4/7-2cos^2/5) +C

Answer 2

追答