To solve the integral \(\int \frac{\sqrt{\tan x}}{\sin^2 x} \, dx\), let's break it down step by step.

### Step 1: Simplify the Integral
First, recognize that \(\tan x = \frac{\sin x}{\cos x}\). Thus, we can rewrite the integral as:

\[
\int \frac{\sqrt{\frac{\sin x}{\cos x}}}{\sin^2 x} \, dx
\]

This simplifies to:

\[
\int \frac{\sqrt{\sin x}}{\sqrt{\cos x} \cdot \sin^2 x} \, dx
\]

We can further simplify this expression:

\[
\int \frac{1}{\sqrt{\cos x} \cdot \sin^{3/2} x} \, dx
\]

### Step 2: Use Substitution
Let's make the substitution \( u = \sin x \), which gives us \( du = \cos x \, dx \). The integral becomes:

\[
\int \frac{1}{\sqrt{1-u^2} \cdot u^{3/2}} \cdot \frac{du}{\cos x}
\]

Since \( \cos x = \sqrt{1 - u^2} \), this simplifies to:

\[
\int \frac{1}{u^{3/2}} \, du
\]

### Step 3: Integrate
Now, integrate the expression:

\[
\int u^{-3/2} \, du
\]

This is a standard integral, which results in:

\[
\frac{u^{-1/2}}{-1/2} = -2u^{-1/2} + C
\]

### Step 4: Substitute Back
Finally, substitute back \( u = \sin x \) to get the final answer:

\[
-2 \frac{1}{\sqrt{\sin x}} + C
\]

So, the integral is:

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

Would you like further details on any step? Here are five related questions for you:

1. How does the substitution method work in integrals?
2. What are other standard trigonometric identities used in integrals?
3. How do you approach integrals involving trigonometric functions and roots?
4. What is the significance of the constant of integration, \( C \)?
5. How would you evaluate definite integrals involving similar functions?

**Tip:** When dealing with integrals involving trigonometric functions, substitution can often simplify the problem significantly.

Explore the step-by-step solution to the integral ∫ sqrt(tan(x))/sin^2(x) dx. Learn how to apply the substitution method and trigonometric identities to simplify and integrate the expression effectively. Suitable for undergraduate students and anyone interested in advanced calculus problems.

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