∫（1/cosX）dX这个不定积分等于?
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cos2x = 2(cosx)^2 -1 = 2/(secx)^2 - 1 = 2/[1+(tanx)^2] - 1 = [1-(tanx)^2]/[1+(tanx)^2]
设 t = tan(x/2),x = 2*arctan(t),dt = [sec(x/2)]^2 * 1/2*dx = 1/2*(1+t^2)*dx,dx = 2dt/(1+t^2)
∫dx/cosx
=∫dx *{1+[tan(x/2)]^2}/{1-[tan(x/2)]^2}
=∫2dt/(1+t^2) * (1+t^2) /(1-t^2)
=2∫dt/(1-t^2)
=∫[1/(1-t) - 1/(t+1)]*dt
=∫dt/(1-t) - ∫dt/(1+t)
=-ln|1-t| - ln|1+t| + C
=-ln|1-t^2| + C
=-ln|1-[tan(x/2)]^2| + C
设 t = tan(x/2),x = 2*arctan(t),dt = [sec(x/2)]^2 * 1/2*dx = 1/2*(1+t^2)*dx,dx = 2dt/(1+t^2)
∫dx/cosx
=∫dx *{1+[tan(x/2)]^2}/{1-[tan(x/2)]^2}
=∫2dt/(1+t^2) * (1+t^2) /(1-t^2)
=2∫dt/(1-t^2)
=∫[1/(1-t) - 1/(t+1)]*dt
=∫dt/(1-t) - ∫dt/(1+t)
=-ln|1-t| - ln|1+t| + C
=-ln|1-t^2| + C
=-ln|1-[tan(x/2)]^2| + C
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