What is $\int \frac{{sec}^{2} (x)}{\sqrt{1 - {tan}^{2} x}} d x$ $int sec^2(x)/(sqrt(1-tan^2x)) dx$ ?

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Answer 1 · 2018-03-22T07:57:59

${sin}^{- 1} (tan x) + c$ $sin^(-1)(tanx)+c$

$I = \int (\frac{{sec}^{2} x}{\sqrt{1 - {tan}^{2} x}}) d x$ $I=int(sec^2x/(sqrt(1-tan^2x)))dx$

us ethe substitution

$u = tan x$ $u=tanx$

$:. du=sec^2xdx$

$=>I=int((sec^2x)/sqrt(1-u^2))xx (du)/sec^2x$

$I=int (du)/sqrt(1-u^2)$

this is a standard integral

$I=sin^(-1)(u)+c$

$=sin^(-1)(tanx)+c$

What is int sec^2(x)/(sqrt(1-tan^2x)) dx∫sec2(x)√1−tan2xdxint sec^2(x)/(sqrt(1-tan^2x)) dx?