To solve the integral $\int_{0}^{\frac{\pi}{4}} \sin(2x) \, dx$, we'll follow these steps:

Step 1: Identify the integral

The integral is $\int \sin(2x) \, dx$. The limits of integration are from $0$ to $\frac{\pi}{4}$.

Step 2: Perform the substitution

To solve the integral, we use the substitution method. Let $u = 2x$. Then, $du = 2 \, dx$, or $dx = \frac{1}{2} \, du$.

Substituting into the integral: $\int \sin(2x) \, dx = \int \sin(u) \cdot \frac{1}{2} \, du = \frac{1}{2} \int \sin(u) \, du$

Step 3: Integrate

The integral of $\sin(u)$ is $-\cos(u)$, so: $\frac{1}{2} \int \sin(u) \, du = -\frac{1}{2} \cos(u) + C$

Step 4: Substitute back the original variable

Since $u = 2x$, substitute back: $-\frac{1}{2} \cos(2x) + C$

Step 5: Apply the limits of integration

Now, we evaluate the definite integral from $0$ to $\frac{\pi}{4}$: $\int_{0}^{\frac{\pi}{4}} \sin(2x) \, dx = -\frac{1}{2} \left[ \cos(2x) \right]_{0}^{\frac{\pi}{4}}$ This evaluates to: $-\frac{1}{2} \left( \cos\left(\frac{\pi}{2}\right) - \cos(0) \right)$ $= -\frac{1}{2} \left( 0 - 1 \right)$ $= \frac{1}{2}$

Final Answer:

The value of the integral $\int_{0}^{\frac{\pi}{4}} \sin(2x) \, dx$ is $\frac{1}{2}$.

Would you like further details or have any questions?

Here are 8 questions related to the problem:

1. How would the integral change if the limits were $0$ to $\frac{\pi}{2}$?
2. What is the integral of $\cos(2x)$ with respect to $x$?
3. How would you approach the integration of $\sin^2(x)$?
4. What is the significance of substitution in solving integrals?
5. How can you verify the result of this integral using numerical methods?
6. What is the physical interpretation of an integral in a real-world context?
7. How would the integration process differ if the function were $\sin^3(2x)$?
8. How does the fundamental theorem of calculus apply to definite integrals?

Tip: Understanding substitution is key in solving integrals involving trigonometric functions. Practice with different trigonometric identities to improve your integration skills.