To solve the given integral:

$\int_{\pi/3}^{\pi/2} \frac{\sin x}{\sqrt{1 - 2 \cos x}} \, dx$

we will follow these steps:

Step 1: Substitution

Let $u = 1 - 2\cos x$, so $du = 2\sin x \, dx$. This substitution simplifies the denominator and involves the sine term in the numerator.

Step 2: Adjusting Limits

Under the substitution $u = 1 - 2\cos x$:

• When $x = \pi/3$, $\cos(\pi/3) = 1/2$, so $u = 1 - 2(1/2) = 0$.
• When $x = \pi/2$, $\cos(\pi/2) = 0$, so $u = 1 - 2(0) = 1$.

Thus, the integral becomes: $\int_{0}^{1} \frac{1}{\sqrt{u}} \cdot \frac{du}{2}.$

Step 3: Simplify the Integral

$\int_{0}^{1} \frac{1}{\sqrt{u}} \cdot \frac{1}{2} \, du = \frac{1}{2} \int_{0}^{1} u^{-1/2} \, du.$

The integral of $u^{-1/2}$ is: $\int u^{-1/2} \, du = 2u^{1/2}.$

So: $\frac{1}{2} \int_{0}^{1} u^{-1/2} \, du = \frac{1}{2} \cdot [2u^{1/2}]_{0}^{1}.$

Step 4: Evaluate the Limits

$\frac{1}{2} \cdot [2u^{1/2}]_{0}^{1} = \frac{1}{2} \cdot (2(1)^{1/2} - 2(0)^{1/2}) = \frac{1}{2} \cdot 2 = 1.$

Final Answer:

The value of the integral is: $\boxed{1}.$

Would you like further explanation of any step?

Here are some related questions:

1. How does substitution simplify integrals involving trigonometric functions?
2. Why do we adjust the limits when using substitution in definite integrals?
3. What are some common trigonometric identities useful in integration?
4. How can you verify the result of a definite integral numerically?
5. What other methods can be used to evaluate integrals with square root terms?

Tip: Always re-check substitutions and limits in definite integrals to avoid errors in the final result.