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Mathematics Question on Integrals of Some Particular Functions

∫sec x dxcos 2 x\int \frac{sec\,x\,dx}{\sqrt{cos\,2\,x}} is equal to

A

2 sin⁡−1 (tan⁡ x)+C2\, \sin^{-1}\, (\tan\,x) + C

B

tan⁡−1 (tan⁡ x2)+C\tan^{-1}\,\left(\frac{\tan\,x}{2}\right)+C

C

sin⁡−1 (tan⁡ x)+C\sin^{-1}\, (\tan\,x) + C

D

12sin⁡−1 (tan⁡ x)+C\frac{1}{2}\sin^{-1}\,\left(\tan\,x\right)+C

Answer

sin⁡−1 (tan⁡ x)+C\sin^{-1}\, (\tan\,x) + C

Explanation

Solution

Let I=∫sec⁡xcos⁡2xdx I =\int \frac{\sec x}{\sqrt{\cos 2 x}} d x
=∫sec⁡x1−tan⁡2x1+tan⁡2xdx=\int \frac{\sec x}{\sqrt{\frac{1-\tan ^{2} x}{1+\tan ^{2} x}}} d x
=∫sec⁡2x1−tan⁡2xdx=\int \frac{\sec ^{2} x}{\sqrt{1-\tan ^{2} x}} d x
Put tan⁡x=t\tan x=t
⇒sec⁡2xdx=dt\Rightarrow \sec ^{2} x d x =d t
∴I=∫dt1−t2\therefore I =\int \frac{d t}{\sqrt{1-t^{2}}}
=sin⁡−1t+C=\sin ^{-1} t+C
=sin⁡−1(tan⁡x)+C=\sin ^{-1}(\tan x)+C