What is the integral of (3sin^2(θ/2)+1/2) ?

The integral of (3sin^2(θ/2)+1/2) is 2θ-3/2 sin(θ)+C

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integral of (3sin^2(θ/2)+1/2)

\int (3 sin^{2} (\frac{θ}{2}) + \frac{1}{2}) d θ

2 θ - \frac{3}{2} sin (θ) + C

Solution steps

\int 3 sin^{2} (\frac{θ}{2}) + \frac{1}{2} 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

= \int 3 sin^{2} (\frac{θ}{2}) d θ + \int \frac{1}{2} d θ

\int 3 sin^{2} (\frac{θ}{2}) d θ = \frac{3}{2} (θ - sin (θ))

\int \frac{1}{2} d θ = \frac{1}{2} θ

= \frac{3}{2} (θ - sin (θ)) + \frac{1}{2} θ

Simplify

\frac{3}{2} (θ - sin (θ)) + \frac{1}{2} θ : 2 θ - \frac{3}{2} sin (θ)

= 2 θ - \frac{3}{2} sin (θ)

Add a constant to the solution

= 2 θ - \frac{3}{2} sin (θ) + C

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What is the integral of (3sin^2(θ/2)+1/2) ?
The integral of (3sin^2(θ/2)+1/2) is 2θ-3/2 sin(θ)+C