Question

\(\int (\sqrt{\tan x} + \sqrt{\cot x}) \, dx =\)

• A. \(2 \tan^{-1}\left( \frac{\tan x - 1}{\sqrt{2 \tan x}} \right) + c\)
• B. \(\tan^{-1}\left( \frac{\tan x - 2}{2 \sqrt{\tan x}} \right) + c\)
• C. \(\sqrt{2} \tan^{-1}\left( \frac{\tan x - 1}{\sqrt{2 \tan x}} \right) + c\)
• D. \(\sqrt{2} \tan^{-1}\left( \frac{\tan x + 1}{\sqrt{2 \tan x}} \right) + c\)

Hint:

For integrals involving \(\sqrt{\tan x}\), use substitution \(u = \sqrt{\tan x}\). Factorize denominators like \(u^4 + 1\) for partial fractions.

Answer 1

The Correct Option is C

Rewrite the integrand: \[ \sqrt{\tan x} + \sqrt{\cot x} = \frac{\sqrt{\tan x} \cdot \sqrt{\tan x} + \sqrt{\cot x} \cdot \sqrt{\tan x}}{\sqrt{\tan x}} = \frac{\tan x + 1}{\sqrt{\tan x}} \] Let \(u = \sqrt{\tan x}\), so \(u^2 = \tan x\), and: \[ \sec^2 x \, dx = 2u \, du \implies dx = \frac{2u}{1 + u^4} \, du \] The integral becomes: \[ \int \frac{u^2 + 1}{u} \cdot \frac{2u}{1 + u^4} \, du = \int \frac{2 (u^2 + 1)}{1 + u^4} \, du = 2 \int \frac{u^2 + 1}{u^4 + 1} \, du \] Factor the denominator: \[ u^4 + 1 = (u^2 - \sqrt{2} u + 1)(u^2 + \sqrt{2} u + 1) \] Use partial fractions: \[ \frac{u^2 + 1}{(u^2 - \sqrt{2} u + 1)(u^2 + \sqrt{2} u + 1)} = \frac{A u + B}{u^2 - \sqrt{2} u + 1} + \frac{C u + D}{u^2 + \sqrt{2} u + 1} \] Solving, we find \(A = \frac{\sqrt{2}}{2}\), \(B = \frac{1}{2}\), \(C = -\frac{\sqrt{2}}{2}\), \(D = \frac{1}{2}\). Thus: \[ 2 \int \left( \frac{\frac{\sqrt{2}}{2} u + \frac{1}{2}}{u^2 - \sqrt{2} u + 1} + \frac{-\frac{\sqrt{2}}{2} u + \frac{1}{2}}{u^2 + \sqrt{2} u + 1} \right) \, du \] Integrate each term: \[ \int \frac{\frac{\sqrt{2}}{2} u + \frac{1}{2}}{u^2 - \sqrt{2} u + 1} \, du \] Complete the square: \(u^2 - \sqrt{2} u + 1 = \left( u - \frac{\sqrt{2}}{2} \right)^2 + \frac{1}{2}\). Substitute \(v = u - \frac{\sqrt{2}}{2}\), adjust, and integrate to get terms involving \(\tan^{-1}\). After combining, the result simplifies to: \[ \sqrt{2} \tan^{-1}\left( \frac{u^2 - 1}{\sqrt{2} u} \right) + c \] Since \(u = \sqrt{\tan x}\), \(u^2 = \tan x\): \[ \frac{u^2 - 1}{\sqrt{2} u} = \frac{\tan x - 1}{\sqrt{2 \tan x}} \] \[ I = \sqrt{2} \tan^{-1}\left( \frac{\tan x - 1}{\sqrt{2 \tan x}} \right) + c \] Option (3) is correct. Options (1), (2), and (4) do not match the derived form.