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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!

Evaluate the following integral using U-substitution: ∫[0,π/4] (sin(2x) / (1 + cos(2x))) dx

Learn how to evaluate the integral of sin(2x) / (1 + cos(2x)) using U-substitution. This step-by-step solution involves key concepts of integration and trigonometric identities, suitable for students studying Calculus II or higher.

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