The correct option is A tan x+c ∫1cos2x+sin^2xdx =∫1(cos^2x sin^2x)+sin^2xdx =∫1cos^2xdx =∫^2xdx =tan x+c ...

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Sum of Trigonometric Ratios in Terms of Their Product

∫1/cos2x+sin2...

Question

$\int \frac{1}{cos 2 x + {sin}^{2} x} d x =$

tan x + c

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cot x + c

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\frac{{tan}^{2} x}{2} + c

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\frac{{cot}^{2} x}{2} + c

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Solution

The correct option is A $tan x + c$
$\int \frac{1}{cos 2 x + {sin}^{2} x} d x$
$= \int \frac{1}{({cos}^{2} x - {sin}^{2} x) + {sin}^{2} x} d x$
$= \int \frac{1}{{cos}^{2} x} d x$
$= \int {sec}^{2} x d x$
$= tan x + c$

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\int \frac{\cos 2 x - 1}{\cos 2 x + 1} d x =

is equal to

A. tan x + cot x + C

B. tan x + cosec x + C

C. − tan x + cot x + C

D. tan x + sec x + C

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