Integrate : ∫(√tan x + √cot x) dx, for x ∈ [0,π/2]

1.	Integrate : ∫(√tan x + √cot x) dx, for x ∈ [0,π/2]
Answer» ∫(√tan x + √cot x) dx, for x ∈ [0,π/2] ∫(√(sin x/cos x) + √(cos x/sin x)) dx, for x ∈ [0,π/2] ∫(sin x + cos x)/√(sin x.cos x) dx, for x ∈ [0,π/2] √2∫(sin x + cos x)/√(2sin x.cos x) dx, for x ∈ [0,π/2] √2∫(sin x + cos x)/√(1 - (1 - 2sin x.cos x)) dx, for x ∈ [0,π/2] √2∫(sin x + cos x)/√(1 - (sin x - cos x)²) dx, for x ∈ [0,π/2] Put sin x - cos x = t (cos x + sin x) dx = dt When x = 0, t = 1 and when x = π/2, t = 1 So, ∫(√tan x + √cot x) dx, for x ∈ [0,π/2] √2∫(dt/√(1 - t²), for t ∈ [-1,1] = √2.[sin^-1 t], for t ∈ [-1,1] = √2[sin^-11 - sin^-1(-1)] = √2((π/2) - (-π/2)) = √2.π

Answer»

∫(√tan x + √cot x) dx, for x ∈ [0,π/2]

∫(√(sin x/cos x) + √(cos x/sin x)) dx, for x ∈ [0,π/2]

∫(sin x + cos x)/√(sin x.cos x) dx, for x ∈ [0,π/2]

√2∫(sin x + cos x)/√(2sin x.cos x) dx, for x ∈ [0,π/2]

√2∫(sin x + cos x)/√(1 - (1 - 2sin x.cos x)) dx, for x ∈ [0,π/2]

√2∫(sin x + cos x)/√(1 - (sin x - cos x)²) dx, for x ∈ [0,π/2]

Put sin x - cos x = t

(cos x + sin x) dx = dt

When x = 0, t = 1 and when x = π/2, t = 1

So, ∫(√tan x + √cot x) dx, for x ∈ [0,π/2]

√2∫(dt/√(1 - t²), for t ∈ [-1,1]

= √2.[sin^-1 t], for t ∈ [-1,1]

= √2[sin^-11 - sin^-1(-1)] = √2((π/2) - (-π/2))

= √2.π

Discussion