∫sec^2x/√{16 + tan^2x}dx = ?

Question

\(\int\frac{sec^2x}{\sqrt{16+tan^2x}}dx\) = ?

∫sec²x/√{16 + tan²x}dx

A. log |tan x + \(\sqrt{tan^2x+16}\) +c

B. log |x + \(\sqrt{tan^2x+16}\) +c

C. log |tan x - \(\sqrt{tan^2x+16}\) +c

D. none of these

Answer 1

\(\int\frac{sec^2xdx}{\sqrt{(tanx)^2+(4)^2}}\)

Put t = tan x

dt = sec²x