The correct option is C tan x +c ∫ e^logsec^2 x dx=∫sec^2 x dx #160; #160; #160; #160; #160; #160; #160; #160; #160; # ...

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Integrating Factor

∫ e log sec 2...

Question

$\int e^{l o g {s e c}^{2} x} d x = . . . . . .$

{sec}^{2} x + c

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tan x + c

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cot x + c

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e^{l o g} {s e c}^{2 x} . tan x + c

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Solution

The correct option is C $tan x + c$
$\int e^{log {sec}^{2} x} d x = \int {sec}^{2} x d x$
$= tan x + C$

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