The integral of sin 2x is -½cos(2x) + C, where C is the constant of integration. This result follows from the standard integral of sin(u) and substitution.
To find ∫sin(2x) dx, use substitution:
Substitute into the integral:
∫sin(2x) dx = ∫sin(u) (du/2) = (1/2)∫sin(u) duThe antiderivative of sin(u) is -cos(u), so:
(1/2)∫sin(u) du = (1/2)(-cos(u)) + C = -½cos(2x) + CIntegrals of sin(2x) appear in three common forms:
| Type | Integral | Result |
|---|---|---|
| Basic indefinite integral | ∫sin(2x) dx | -½cos(2x) + C |
| Definite integral /2 | ∫₀ᵖⁱˣ/² sin(2x) dx | -½cos(π) - (-½cos(0)) = ½ |
| Integral multiplied by coefficient | ∫3sin(2x) dx | 3(-½cos(2x)) + C = -½cos(2x) + C |
The integral of sin(2x) appears in physics when calculating work done by oscillating forces. For example, integrating sin(2x) over one period gives zero, showing no net work is done by a force oscillating sinusoidally over complete cycles.
In probability, it appears in the PDF of certain distributions. The chi-squared distribution's PDF contains a term resembling sin(2x)'s integral.
Many students mistakenly think ∫sin(2x) dx = -sin(2x) + C. This error occurs by forgetting the substitution step. The correct antiderivative is always -½cos(2x) + C.
Another frequent mistake is omitting the ½ factor when integrating sin(2x). Always remember to divide by the derivative of the inner function (2 in this case).
Find ∫sin(2x) dx from x=0 to x=π/4.
Using our antiderivative:
∫₀ᵖⁱˣ/⁴ sin(2x) dx = [-½cos(2x)]₀ᵖⁱˣ/⁴ = -½cos(π/2) - (-½cos(0)) = 0 - (-½) = ½This result shows the area under sin(2x) /4 equals ½.
I once used this integral to calculate the average power output of an alternating current system. By integrating sin(2x) over one cycle, we showed the average power was zero, confirming the system's energy oscillations were symmetric.