问题标题:
sinx-1次方的不定积分是多少
问题描述:
sinx-1次方的不定积分是多少
金通回答:
=积分dx/sinx
=积分(sin^2x+cos^2x)dx/sinx,^2表示平方
=积分sinxdx+积分cos^2xdx/sinx
=-cosx+积分cos^2xsinxdx/sin^2x
=-cosx+积分cos^2xsinxdx/(1-cos^2x)
第二个令t=cosx
dt=-sinxdx
=-cosx-积分t^2dt/(1-t^2)
=-cosx+积分dt+积分dt/(t^2-1)
=-cosx+t+(1/2)积分dt[1/(t-1)-1/(t+1)]
=-cosx+cosx+(1/2)积分[dt/(t-1)-dt/(t+1)]
=(1/2)(ln|t-1|-ln|t+1|)+C
=(1/2)ln|(cosx-1)/(cosx+1)|+C
代入cosx=(1-tan^2(x/2))/(1+tan^2(x/2))
=(1/2)ln|tan^2(x/2)|+C
=ln|tan(x/2)|+C
=ln|(1-cosx)/sinx|+C
=ln|cscx-cotx|+C
点击显示
数学推荐
热门数学推荐