# Integration of sec²x / 1-tan²x ds

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$\int\frac{sec^x}{1-tan^2x}dx$

= $\int\frac{dt}{1-t^2}$ ($\because t = tan x$, ⇒ sec2x dx = dt)

= $\frac12log|\frac{1+t}{1-t}|+c$

= $\frac12log|\frac{1+tan x}{1-tanx}|+c$

= $\frac12log|tan(\frac {\pi}4+x)|+c$