Math Problem Statement

integral sin(4x)dx

Solution

The integral of sin(4x)\sin(4x) with respect to xx can be solved using basic integration rules. Here's the solution:

sin(4x)dx\int \sin(4x) \, dx

To solve this, we use substitution. Let:

u=4xwhich meansdu=4dxordx=du4u = 4x \quad \text{which means} \quad du = 4 \, dx \quad \text{or} \quad dx = \frac{du}{4}

Now substitute uu into the integral:

sin(4x)dx=sin(u)du4\int \sin(4x) \, dx = \int \sin(u) \cdot \frac{du}{4}

Factor out the constant 14\frac{1}{4}:

14sin(u)du\frac{1}{4} \int \sin(u) \, du

The integral of sin(u)\sin(u) is cos(u)-\cos(u), so:

14(cos(u))+C\frac{1}{4} \cdot (-\cos(u)) + C

Substitute u=4xu = 4x back into the equation:

14cos(4x)+C-\frac{1}{4} \cos(4x) + C

Thus, the integral is:

sin(4x)dx=14cos(4x)+C\int \sin(4x) \, dx = -\frac{1}{4} \cos(4x) + C

Would you like more details on the steps or have any questions?

Here are 5 related questions:

  1. How do you integrate functions with other trigonometric identities, such as cos(5x)\cos(5x)?
  2. What is the integral of a product of sine and cosine, such as sin(x)cos(x)\sin(x)\cos(x)?
  3. Can substitution be used to solve integrals of higher powers of trigonometric functions?
  4. What is the derivative of 14cos(4x)-\frac{1}{4} \cos(4x)?
  5. How does the integral of sin(4x)\sin(4x) compare to sin(kx)\sin(kx), where kk is a constant?

Tip: Always look for substitutions in integrals involving complex trigonometric functions; it often simplifies the process significantly.

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Math Problem Analysis

Mathematical Concepts

Calculus
Trigonometric Integration
Substitution

Formulas

∫sin(kx) dx = -(1/k) cos(kx) + C
Substitution: u = 4x, du = 4 dx

Theorems

Basic Integration Rules

Suitable Grade Level

Grades 11-12, College