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{\partial}{\partial s} (4 s t)

\frac{d y}{d x} = y - 2 x

x \to 4 lim (x - 6)

x \cdot \frac{d y}{d x} + 3 y = x^{3} - x

f (x) = (6 x^{2} + 4 x)^{2}

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