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
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