Question

∫[dx/√(1 + cosec²x)] = _________ + c

(a) sin^–1[(sinx)/(√2)]

(b) sin^–1[(cosx)/(√2)]

(c) cos^–1[(cosx)/√2]

(d) cos^–1[(sinx)/√2]

Answer 1

The correct option (b) sin^–1[(cosx)/(√2)]   

Explanation:

I = ∫[dx / √(1 + cosec²x)] = ∫[(sinx)/√(sin²x + 1)]dx

∴ I = ∫[(sinx)/√(2 – cos²x)]dx

Let    cosx = t ⇒ – sinx dx = dt

∴    I = ∫– [(dt) / √(2 – t²)]

= ∫[(dt) / √(t² – 2)]

= sin^–1(t / √2) + c

= sin^–1[(cosx) / √2] + c