∫(secx/tan^2x)dx
=積分:1/(sinx*tanx)dx=積分:cosx/sin^2xdx
設(shè)tanx/2=t,
sinx=2t/(1+t^2)
cosx=(1-t^2)/(1+t^2)
dx=2dt/(1+t^2)
=積分:(1-t^2)/(1+t^2)/4t^2/(1+t^2)^2*2dt/(1+t^2)
=積分:(1-4t)/4t^2dt
=積分:1/4t^2-1/tdt
=-1/4t-ln|t|+C
將t=tanx/2代入 就可以了
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∫[secx/(tanx)^2]dx
因?yàn)閟ecx=1/cosx
所以∫[secx/(tanx)^2]dx
=∫[1/cosxtan^2xdx
=積分:1/sinxtanxdx
=積分:cosx/sin^2xdx
=∫[1/(sinx)^2]d(sinx)
=-1/(sinx)