Math Problem Statement

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Solution

To solve the integral 0π/4sin(2x)dx\int_{0}^{\pi/4} \sin(2x) \, dx, we will follow these steps:

Step-by-Step Solution

  1. Identify the Integral: sin(2x)dx\int \sin(2x) \, dx

  2. Substitution: To solve this integral, use a substitution method. Let u=2xu = 2x. Then, du=2dxdu = 2 \, dx or dx=12dudx = \frac{1}{2} \, du.

  3. Change Limits of Integration: When x=0x = 0, u=2×0=0u = 2 \times 0 = 0.

    When x=π4x = \frac{\pi}{4}, u=2×π4=π2u = 2 \times \frac{\pi}{4} = \frac{\pi}{2}.

    The integral becomes: 0π/4sin(2x)dx=120π/2sin(u)du\int_{0}^{\pi/4} \sin(2x) \, dx = \frac{1}{2} \int_{0}^{\pi/2} \sin(u) \, du

  4. Evaluate the Integral: 120π/2sin(u)du=12[cos(u)]0π/2\frac{1}{2} \int_{0}^{\pi/2} \sin(u) \, du = \frac{1}{2} [-\cos(u)]_{0}^{\pi/2}

  5. Calculate the Definite Integral: =12(cos(π2)+cos(0))= \frac{1}{2} \left( -\cos\left(\frac{\pi}{2}\right) + \cos(0) \right)

    =12(0+1)=12×1=12= \frac{1}{2} \left( -0 + 1 \right) = \frac{1}{2} \times 1 = \frac{1}{2}

Final Answer

0π/4sin(2x)dx=12\int_{0}^{\pi/4} \sin(2x) \, dx = \frac{1}{2}

Let me know if you have any questions or would like further clarification on any part of this process.

Related Questions

  1. How do you solve integrals using substitution?
  2. What is the geometric interpretation of definite integrals?
  3. How do the limits of integration change during substitution?
  4. What are some common trigonometric identities used in integration?
  5. Can you solve cos(2x)dx\int \cos(2x) \, dx using a similar method?
  6. What is the difference between definite and indefinite integrals?
  7. How can you verify the result of a definite integral?
  8. What other integration techniques are there besides substitution?

Tip

When using substitution in integrals, always remember to change the limits of integration to match the new variable to simplify the process.

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Math Problem Analysis

Mathematical Concepts

Calculus
Trigonometry
Integration

Formulas

Substitution method for integration

Theorems

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Suitable Grade Level

Advanced High School and Beyond