Question

求〔sin(x的三次方)cosx〕的不定积分

Answer 1

∫sinxdx/(sinx^3+cosx^3) =∫dx/sinx^2(1+cotx^3) =-∫dcotx/(1+cotx^3) cotx=u =-∫du/(1+u^3) =(-1/6)ln|u^2-u+1|+(1/√3)arctan[(2u-1)/√3] +(1/3)ln|u+1|+C =(-1/6)ln|cotx^2-cotx+1| +(1/√3)arctan[(2cotx-1)/√3]+(1/√3ln|cotx+1|+C ∫dx/(1+x^3)=∫dx/[(1+x)(1+x^2-x)]=(1/3)∫(x+1)^2-(x^2-x+1)dx/[x(1+x)(1-x+x^2)] =(1/3)∫(x+1)dx/x(1-x+x^2)-(1/3)∫dx/x(1+x) =(1/3)∫dx/x(1-x+x^2)+ (1/3)∫dx/(1-x+x^2)+(1/3)ln(1+x)/x =(1/6)∫dx^2/x^2(1-x+x^2)+... =(1/6)∫(x-1)dx^2/[x^2(1-x+x^2)(x-1)]+.. =(1/6)∫dx^2/(x-1)(1-x+x^2)-(1/6)∫dx^2/x^2(x-1)+... =(1/3)∫dx/(x-1)(1-x+x^2)-(1/3)∫dx/x(x-1)+... =(1/3)∫dx/(x-1)-(1/3)∫(x-1)dx/(1-x+x^2)-(1/3)ln(x-1)/x+... =(1/3)ln(x-1)-(1/6)∫d(x^2-x+1)/(x^2-x+1)+(1/6)∫dx/(1-x+x^2)-(1/3)ln(x-1)/x+... =(-1/6)ln|x^2-x+1|+(1/√3)arctan[(2x-1)/√3] +(1/3)ln|x+1|+C