Question

$integrate \\ \frac{sec \: x}{ \sqrt{cos \: 2x} } \: dx$

(a) tan-1(cos x)+c
(c) tan -1(sin x) +c
(b) sin-1(tan x) +c
(d) 2 sin -1(tan x)+e​

Answer 1

Answer:

$\large\boxed{\sf{ (b)\:{ \sin }^{ - 1} ( \tan x) + c}}$

Step-by-step explanation:

$\displaystyle \int \frac{ \sec(x) }{ \sqrt{ \cos(2x) } } dx \\ \\ = \displaystyle \int \frac{ \sec(x) }{ \sqrt{ { \cos }^{2}x - { \sin}^{2}x } }dx \\ \\ = \displaystyle \int \frac{ \sec(x) }{ \cos(x) \sqrt{1 - { \tan }^{2}x } } dx \\ \\ = \displaystyle \int \frac{ { \sec }^{2} x}{ \sqrt{1 - { \tan }^{2} x} } dx$

Let

$\tan(x) = t \\ \\ = > { \sec }^{2}x \: dx = dt$

Thus we get

$= \displaystyle \int \frac{dt}{ \sqrt{1 - {t}^{2} } } \\ \\ = { \sin }^{ - 1} t + c \\ \\ = { \sin }^{ - 1} ( \tan x) + c$