What is the integral of 2sin^2(θ/2) ?

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

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integral of 2sin^2(θ/2)

\int 2 sin^{2} (\frac{θ}{2}) d θ

θ - sin (θ) + C

Solution steps

\int 2 sin^{2} (\frac{θ}{2}) d θ

Take the constant out:

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

= 2 \cdot \int sin^{2} (\frac{θ}{2}) d θ

Rewrite using trig identities

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

Take the constant out:

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

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

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

\int 1 d θ = θ

\int cos (θ) d θ = sin (θ)

= 2 \cdot \frac{1}{2} (θ - sin (θ))

Simplify

2 \cdot \frac{1}{2} (θ - sin (θ)) : θ - sin (θ)

= θ - sin (θ)

Add a constant to the solution

= θ - sin (θ) + C

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