Question

If $\int_{}^{}\sqrt{1 + \sec x}$dx = K sin^–1 (ƒ(x)) + c then –

• A. ƒ(x) = $\sqrt{2}$ sin (x/ 2)
• B. ƒ(x) = $\sqrt{2}$cos (x/2), K= 2
• C. ƒ(x) = $\sqrt{2}$tan (x/2), K = 2
• D. ƒ(x) =$\sqrt{2}$sin (x/2), K =$\sqrt{2}$

Answer 1

Answer: (A)

ƒ(x) = $\sqrt{2}$ sin (x/ 2)

Answer 2

$\sqrt{1 + \sec x}$dx= $\sqrt{\frac{1 + \cos x}{\cos x}}$= $\frac{\sqrt{2}\cos x/2}{\sqrt{1 - 2\sin^{2}x/2}}$so putting

sin x/2 = t, we have

$\int_{}^{}\sqrt{1 + \sec x}$dx= $\int_{}^{}\frac{\sqrt{2}\cos x/2}{\sqrt{1 - 2\sin^{2}x/2}}$dx

= 2 sin^–1 $\sqrt{2}$t + c = 2 sin^–1 ($\sqrt{2}$ sin x/2) + c

So, ƒ(x) = $\sqrt{2}$sin x/2 and K = 2