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Question: If \(\int_{}^{}\sqrt{1 + \sec x}\)dx = 2 (ƒog) (x) + C, then –...

If ∫1+sec⁡x\int_{}^{}\sqrt{1 + \sec x}dx = 2 (ƒog) (x) + C, then –

A

ƒ(x) = sec x – 1

B

ƒ(x) = 2 tan–1x

C

ƒ(x) = sec⁡x−1\sqrt{\sec x - 1}q

D

None of these

Answer

ƒ(x) = sec⁡x−1\sqrt{\sec x - 1}q

Explanation

Solution

∫1+sec⁡x\int \sqrt { 1 + \sec x }dx = ∫sec⁡x.tan⁡xsec⁡x.sec⁡x−1\int_{}^{}\frac{\sec x.\tan x}{\sec x.\sqrt{\sec x - 1}}dx

= ∫tan⁡xsec⁡x−1\int \frac { \tan x } { \sec x - 1 } dx

(Put sec x = t Ž sec x tan x dx = dt)

=

= 2 (where y2 = t – 1)

= 2 tan–1 y + C = 2 tan–1 sec⁡x–1\sqrt{\sec x–1} + C

\ ƒ(x) = tan–1 x and g(x) = sec⁡x–1\sqrt{\sec x–1}.