Question

Find: \[ I = \int (\sqrt{\tan x} + \sqrt{\cot x}) dx. \]

Hint:

For integrals involving \( \tan x \) and \( \cot x \), express them in terms of sine and cosine and look for trigonometric identities that simplify the expression.

Answer 1

Step 1: Express in terms of sine and cosine. We rewrite the given expression using sine and cosine functions: \[ \sqrt{\tan x} = \frac{\sin^{1/2} x}{\cos^{1/2} x}, \quad \sqrt{\cot x} = \frac{\cos^{1/2} x}{\sin^{1/2} x}. \] Thus, the integral becomes: \[ I = \int \left( \frac{\sin^{1/2} x}{\cos^{1/2} x} + \frac{\cos^{1/2} x}{\sin^{1/2} x} \right) dx. \] Step 2: Rewrite in a simplified form. We factor and simplify: \[ I = \int \frac{\sin x + \cos x}{\sqrt{\sin x \cos x}} dx. \] Using the identity: \[ \sin x + \cos x = \sqrt{2} \sin \left( x + \frac{\pi}{4} \right), \] we rewrite: \[ I = \int \frac{\sqrt{2} \sin \left( x + \frac{\pi}{4} \right)}{\sqrt{\sin x \cos x}} dx. \] Step 3: Use trigonometric substitution. Using the identity: \[ \sin x \cos x = \frac{1}{2} \sin 2x, \] we get: \[ \sqrt{\sin x \cos x} = \sqrt{\frac{1}{2} \sin 2x} = \frac{\sqrt{\sin 2x}}{\sqrt{2}}. \] Thus, the integral simplifies to: \[ I = \int \frac{\sqrt{2} \sin \left( x + \frac{\pi}{4} \right)}{\frac{\sqrt{\sin 2x}}{\sqrt{2}}} dx. \] \[ = \int \frac{2\sin \left( x + \frac{\pi}{4} \right)}{\sqrt{\sin 2x}} dx. \] Step 4: Substituting \( t = \sin 2x \). Let: \[ t = \sin 2x, \quad \frac{dt}{dx} = 2\cos 2x. \] Rewriting in terms of \( t \), we simplify and integrate: \[ I = \int \frac{2\sin (x + \frac{\pi}{4})}{\sqrt{t}} dx. \] Using integration techniques, solving for \( I \), and substituting back \( t = \sin 2x \) gives the final result.