We are given the integral:

$\int_0^{\frac{\pi}{4}} \frac{\sin(2x)}{1 + \cos(2x)} \, dx$

To evaluate this using $u$-substitution, let's follow the steps below:

Step 1: Identify the substitution

Notice that the derivative of $\cos(2x)$ is related to $\sin(2x)$, so we can make the substitution: $u = 1 + \cos(2x)$

Now, differentiate both sides with respect to $x$: $\frac{du}{dx} = -2 \sin(2x)$ or $du = -2 \sin(2x) \, dx$

This suggests that $\sin(2x) \, dx = -\frac{1}{2} \, du$.

Step 2: Change the limits of integration

When $x = 0$: $u = 1 + \cos(0) = 1 + 1 = 2$

When $x = \frac{\pi}{4}$: $u = 1 + \cos\left(2 \cdot \frac{\pi}{4}\right) = 1 + \cos\left(\frac{\pi}{2}\right) = 1 + 0 = 1$

Thus, the limits of integration change from $x = 0$ to $x = \frac{\pi}{4}$ into $u = 2$ to $u = 1$.

Step 3: Rewrite the integral

Substituting everything into the integral, we get: $\int_2^1 \frac{\sin(2x)}{1 + \cos(2x)} \, dx = \int_2^1 \frac{-\frac{1}{2} \, du}{u} = \frac{1}{2} \int_1^2 \frac{du}{u}$

Step 4: Evaluate the integral

The integral of $\frac{1}{u}$ is $\ln|u|$, so we have: $\frac{1}{2} \left[ \ln|u| \right]_1^2 = \frac{1}{2} \left( \ln(2) - \ln(1) \right) = \frac{1}{2} \ln(2)$

Since $\ln(1) = 0$, the final answer is: $\frac{1}{2} \ln(2)$

Final Answer:

$\int_0^{\frac{\pi}{4}} \frac{\sin(2x)}{1 + \cos(2x)} \, dx = \frac{1}{2} \ln(2)$

Would you like further details or explanations? Here are some related questions you might find helpful:

1. How does the choice of $u = 1 + \cos(2x)$ simplify the integral?
2. Why is it important to change the limits of integration after substitution?
3. What are other methods to evaluate trigonometric integrals besides $u$-substitution?
4. How can we use substitution for integrals involving different trigonometric identities?
5. What happens if we substitute $u = \cos(2x)$ instead of $u = 1 + \cos(2x)$?

Tip: When using $u$-substitution, always remember to adjust the limits of integration if the bounds are in terms of the original variable!