I=∫sec⁡xdx1−Tan2⁡x1+Tan2⁡x=∫sec2⁡xdx1−Tan2⁡x Put Tan⁡x=t⇒sec2⁡xdx=dt I=∫dt1-t2=sin-1t+c =sin-1tanx+c

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$\int \frac{dx}{\cos x \sqrt{\cos 2 x}}$

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cos^-1(tanx)+c

cos(tan^-1x)+c

tan(sin^-1x)+c

sin^-1(tanx)+c

answer is C.

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Detailed Solution

$I = \int \frac{\sec xdx}{\sqrt{\frac{1 - {Tan}^{2} x}{1 + {Tan}^{2} x}}} = \int \frac{\sec^{2} xdx}{\sqrt{1 - {Tan}^{2} x}} Put Tan x = t \Rightarrow \sec^{2} xdx = dt I = \int \frac{dt}{\sqrt{1 - t^{2}}} = \sin^{- 1} (t) + c = \sin^{- 1} (tanx) + c$