The given integral is:

$\int_{\frac{1}{3}}^{\frac{1}{2}} \frac{dx}{(1 - x)\sqrt{1 - x^2}}$

Let's solve this step-by-step:

Step 1: Identify the structure of the integrand

We have the expression:

$\frac{1}{(1 - x)\sqrt{1 - x^2}}$

This suggests a potential substitution involving trigonometric identities due to the square root of $1 - x^2$.

Step 2: Trigonometric substitution

Since we recognize $\sqrt{1 - x^2}$, the substitution $x = \sin \theta$ might be useful. Let’s try it:

1. Set $x = \sin \theta$, so $dx = \cos \theta \, d\theta$.
2. Then $\sqrt{1 - x^2} = \sqrt{1 - \sin^2 \theta} = \cos \theta$.

Thus, the integral becomes:

$\int \frac{\cos \theta \, d\theta}{(1 - \sin \theta)\cos \theta}$

Simplifying:

$\int \frac{d\theta}{1 - \sin \theta}$

Step 3: Use of identity for simplification

We use the identity:

$1 - \sin \theta = \frac{(1 - \sin \theta)(1 + \sin \theta)}{1 + \sin \theta} = \frac{1 - \sin^2 \theta}{1 + \sin \theta} = \frac{\cos^2 \theta}{1 + \sin \theta}$

Thus, the integral becomes:

$\int \frac{d\theta}{\frac{\cos^2 \theta}{1 + \sin \theta}} = \int \frac{1 + \sin \theta}{\cos^2 \theta} d\theta$

Now split this into two integrals:

$\int \frac{1}{\cos^2 \theta} d\theta + \int \frac{\sin \theta}{\cos^2 \theta} d\theta$

The first integral is:

$\int \sec^2 \theta \, d\theta = \tan \theta$

The second integral is:

$\int \frac{\sin \theta}{\cos^2 \theta} d\theta = \sec \theta$

Step 4: Combine results

So, the solution for the integral becomes:

$\tan \theta + \sec \theta$

Step 5: Back-substitute $x = \sin \theta$

Recall that $x = \sin \theta$, so:

• $\tan \theta = \frac{x}{\sqrt{1 - x^2}}$
• $\sec \theta = \frac{1}{\sqrt{1 - x^2}}$

Thus, the integral becomes:

$\frac{x}{\sqrt{1 - x^2}} + \frac{1}{\sqrt{1 - x^2}} = \frac{1 + x}{\sqrt{1 - x^2}}$

Step 6: Evaluate the definite integral

We now evaluate the definite integral from $x = \frac{1}{3}$ to $x = \frac{1}{2}$.

First, plug in $x = \frac{1}{2}$:

$\frac{1 + \frac{1}{2}}{\sqrt{1 - \left( \frac{1}{2} \right)^2}} = \frac{\frac{3}{2}}{\sqrt{1 - \frac{1}{4}}} = \frac{\frac{3}{2}}{\frac{\sqrt{3}}{2}} = \frac{3}{\sqrt{3}} = \sqrt{3}$

Now, plug in $x = \frac{1}{3}$:

$\frac{1 + \frac{1}{3}}{\sqrt{1 - \left( \frac{1}{3} \right)^2}} = \frac{\frac{4}{3}}{\sqrt{1 - \frac{1}{9}}} = \frac{\frac{4}{3}}{\frac{\sqrt{8}}{3}} = \frac{4}{\sqrt{8}} = \frac{4}{2\sqrt{2}} = \frac{2}{\sqrt{2}} = \sqrt{2}$

Step 7: Final result

The final value of the definite integral is:

$\sqrt{3} - \sqrt{2}$

Therefore, the value of the given integral is $\sqrt{3} - \sqrt{2}$.

Would you like further details on any step?

Here are 5 related questions to consider:

1. What other substitution methods can simplify integrals with square roots?
2. How does the identity for $1 - \sin \theta$ help in integral evaluation?
3. What are the common scenarios where trigonometric substitution is useful?
4. How do definite integrals with trigonometric substitution work?
5. What are the best practices for checking definite integral results?

Tip: Trigonometric substitution is very effective when dealing with expressions involving square roots like $\sqrt{1 - x^2}$, so identifying these patterns early simplifies integration.