Solving the Integral ∫ √(tan x) / (sin x cos x) dx

Solving the Integral ∫ √(tan x) / (sin x cos x) dx

In this post, we will solve the integral:

\[ \int \frac{\sqrt{\tan x}}{\sin x \cos x} \, dx \]

Step 1: Simplify the Integrand

First, recall the trigonometric identities:

\[ \sin x \cos x = \frac{1}{2} \sin(2x) \] \[ \tan x = \frac{\sin x}{\cos x} \implies \sqrt{\tan x} = \frac{\sqrt{\sin x}}{\sqrt{\cos x}} \]

Rewriting the integrand:

\[ \frac{\sqrt{\tan x}}{\sin x \cos x} = \frac{\sqrt{\sin x}}{\sin x \cos x \sqrt{\cos x}} = \frac{1}{\sqrt{\sin x} \cos^{3/2} x} \]

This simplification doesn't immediately help, so we try a substitution instead.

Step 2: Substitution Approach

Let:

\[ t = \tan x \implies dt = \sec^2 x \, dx \implies dx = \frac{dt}{1 + t^2} \]

Now, express \(\sin x\) and \(\cos x\) in terms of \(t\):

\[ \sin x = \frac{t}{\sqrt{1 + t^2}}, \quad \cos x = \frac{1}{\sqrt{1 + t^2}} \]

Thus, the integrand becomes:

\[ \frac{\sqrt{t}}{\sin x \cos x} = \frac{\sqrt{t}}{\left( \frac{t}{1 + t^2} \right)} = \frac{\sqrt{t} (1 + t^2)}{t} \]

Now, substitute \(dx = \frac{dt}{1 + t^2}\):

\[ \int \frac{\sqrt{t}}{t} \cdot \frac{1 + t^2}{1 + t^2} \, dt = \int \frac{\sqrt{t}}{t} \, dt = \int t^{-1/2} \, dt \]

Step 3: Evaluate the Integral

The integral simplifies to:

\[ \int t^{-1/2} \, dt = 2 \sqrt{t} + C \]

Substitute back \(t = \tan x\):

\[ \boxed{2 \sqrt{\tan x} + C} \]

Final Answer

\[ \int \frac{\sqrt{\tan x}}{\sin x \cos x} \, dx = \boxed{2 \sqrt{\tan x} + C} \]

Conclusion

By using substitution and trigonometric identities, we simplified the integral into a basic power rule problem. The key step was letting \(t = \tan x\), which transformed the integrand into a manageable form.

Let me know in the comments if you have any questions or alternative methods!