2sin3xdx

The integral 2sin(3x)dx is a fundamental problem in calculus, specifically in the field of integration. It is a standard example used to demonstrate techniques for integrating trigonometric functions. The integral can be solved using the substitution method, which is a common technique for integrating functions involving trigonometric identities.

To solve the integral 2sin(3x)dx, we start by recognizing that the derivative of sin(3x) is 3cos(3x). This suggests a substitution where u = 3x, which implies du = 3dx or dx = du/3. Substituting these into the integral, we get:

2sin(3x)dx = 2sin(u) * (du/3)

Now, we can integrate with respect to u:

∫2sin(u) * (du/3) = (2/3) * ∫sin(u)du

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

(2/3) * ∫sin(u)du = (2/3) * (-cos(u)) + C

Substituting back u = 3x, we get the final result:

∫2sin(3x)dx = -(2/3)cos(3x) + C

This result shows that the integral of 2sin(3x) with respect to x is -(2/3)cos(3x) plus a constant of integration C. This solution is a classic example of how trigonometric identities and substitution methods can be used to solve integrals in calculus.