Question

sinx -1次方的不定积分是多少

Answer 1

=积分 dx/sinx
=积分 (sin^2 x +cos^2 x)dx/sinx , ^2表示平方
=积分 sinx dx +积分 cos^2 x dx / sinx
=-cosx +积分 cos^2 x sinx dx /sin^2 x
=-cosx +积分 cos^2 x sinx dx /(1-cos^2 x)
第二个令t=cosx
dt=-sinxdx
=-cosx - 积分 t^2 dt/(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