To solve the given definite integral using U-substitution, let's follow these steps:

Problem

Evaluate the following integral using U-substitution: $\int_0^{\frac{\pi}{4}} \frac{\sin(2x)}{1 + \cos(2x)} \, dx$

Solution

1. Substitute:
   Let $u = 1 + \cos(2x)$.

2. Differentiate:
   Find $du$.
   Since $\frac{d}{dx} \cos(2x) = -2 \sin(2x)$, we have: $du = -2 \sin(2x) \, dx \implies \sin(2x) \, dx = -\frac{1}{2} \, du$

3. Change Limits:
   When $x = 0$:
   $u = 1 + \cos(0) = 2$
   When $x = \frac{\pi}{4}$:
   $u = 1 + \cos\left(2 \times \frac{\pi}{4}\right) = 1 + \cos\left(\frac{\pi}{2}\right) = 1$

4. Rewrite the Integral:
   Substitute into the integral: $\int_{2}^{1} \frac{\sin(2x)}{1 + \cos(2x)} \, dx = \int_{2}^{1} \frac{-\frac{1}{2} \, du}{u}$

5. Simplify and Integrate:
   $= -\frac{1}{2} \int_{2}^{1} \frac{1}{u} \, du = -\frac{1}{2} \left[ \ln |u| \right]_{2}^{1}$

6. Evaluate the Definite Integral: $-\frac{1}{2} \left( \ln(1) - \ln(2) \right) = -\frac{1}{2} (0 - \ln(2)) = \frac{\ln(2)}{2}$

Final Answer

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

Would you like more details or have any questions about the solution?

Related Questions

1. How do you choose the right substitution for an integral?
2. What are the common techniques for integrating trigonometric functions?
3. Can U-substitution be used for indefinite integrals as well?
4. How do you handle integrals that involve squared trigonometric functions?
5. What are some alternative methods for evaluating this integral?

Tip

Always check the new limits of integration after substitution, as they correspond to the new variable $u$.