Integrate ((tanx)^1/2 + (cotx)^1/2)dx

Integration is exactly the reverse process of differentiation. We will use trigonometric formulae to solve this integration

Answer: ∫ ((tanx)^1/2 + (cotx)^1/2)dx = √2 sin^-1(sinx - cosx) + C

Let us solve the problem below

Explanation:

Let, I = ∫ (√tanx + √cotx) dx

I = ∫ (√(sinx/cosx) + √(cosx/sinx)) dx  [Since, tan x = sin x / cos x and cot x = cos x / sin x]

On taking LCM and solving we get,

I = ∫ (sinx + cosx)/√(sinx.cosx) dx ---------------- (1)

Let, sinx - cosx = t 

By differentiating the above we get,

So, (cosx + sinx) dx = dt ---------------- (2)

Also, (sinx - cosx)² = t²

⇒ 1 - 2sinx.cosx  = t²  [Since, sin²x + cos²x = 1]

⇒ sinx.cosx = (1 - t²)/2 ---------------- (3)

Substituting (2) and (3) in (1), we get

I = ∫ √2 dt /√(1 - t²)

Integrating I using the rule, 

∫ dt /√(1 - t²) = sin^-1t

Applying this, we get

I = √2 sin^-1t + C

where, C is the constant of integration.

Therefore, ∫ ((tanx)^1/2 + (cotx)^1/2)dx = √2 sin^-1(sinx - cosx) + C