What is the integral of (1-cos(2θ))/2 ?

The integral of (1-cos(2θ))/2 is 1/2 (θ-1/2 sin(2θ))+C

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integral of (1-cos(2θ))/2

\int \frac{1 - cos ( 2 θ )}{2} d θ

\frac{1}{2} (θ - \frac{1}{2} sin (2 θ)) + C

Solution steps

\int \frac{1 - cos ( 2 θ )}{2} d θ

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{2} \cdot \int 1 - cos (2 θ) d θ

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \frac{1}{2} (\int 1 d θ - \int cos (2 θ) d θ)

\int 1 d θ = θ

\int cos (2 θ) d θ = \frac{1}{2} sin (2 θ)

= \frac{1}{2} (θ - \frac{1}{2} sin (2 θ))

Add a constant to the solution

= \frac{1}{2} (θ - \frac{1}{2} sin (2 θ)) + C

\frac{d}{d x} (2 x^{2} ln (x))

x \to \infty lim (3 - \frac{2}{x})

2 x (\frac{d y}{d x}) = 2 x e^{x} - 2 y + 6 x^{2}

\frac{d}{d x} (A (e^{x} x + 2 e^{x}))

y = \frac{1}{25} x^{2}, x = 5, y = 0

What is the integral of (1-cos(2θ))/2 ?
The integral of (1-cos(2θ))/2 is 1/2 (θ-1/2 sin(2θ))+C