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

The integral of 1/2 (1-cos(θ))^2 is 1/8 (6θ-8sin(θ)+sin(2θ))+C

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integral of 1/2 (1-cos(θ))^2

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

\frac{1}{8} (6 θ - 8 sin (θ) + sin (2 θ)) + C

Solution steps

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

Expand

(1 - cos (θ))^{2} : 1 - 2 cos (θ) + cos^{2} (θ)

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

\int 1 d θ = θ

\int 2 cos (θ) d θ = 2 sin (θ)

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

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

Simplify

\frac{1}{2} (θ - 2 sin (θ) + \frac{1}{2} (θ + \frac{1}{2} sin (2 θ))) : \frac{1}{8} (6 θ - 8 sin (θ) + sin (2 θ))

= \frac{1}{8} (6 θ - 8 sin (θ) + sin (2 θ))

Add a constant to the solution

= \frac{1}{8} (6 θ - 8 sin (θ) + sin (2 θ)) + C

\int (x^{2} - 6 x + 2) \cdot e^{x} d x

\frac{\partial}{\partial x} (3 x^{3} + 2 y^{2})

\frac{4}{x}, \frac{4}{x ^{2}}, 5

\int 2 e^{5 x} d x

n = 0 \sum \infty 2^{n}

What is the integral of 1/2 (1-cos(θ))^2 ?
The integral of 1/2 (1-cos(θ))^2 is 1/8 (6θ-8sin(θ)+sin(2θ))+C