What is the integral of (1-sin^2(x/2)) ?

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

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integral of (1-sin^2(x/2))

\int (1 - sin^{2} (\frac{x}{2})) d x

\frac{1}{2} (x + sin (x)) + C

Solution steps

\int 1 - sin^{2} (\frac{x}{2}) d x

Rewrite using trig identities

= \int \frac{1}{2} (1 + cos (x)) d x

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 (x) d x

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 x + \int cos (x) d x)

\int 1 d x = x

\int cos (x) d x = sin (x)

= \frac{1}{2} (x + sin (x))

Add a constant to the solution

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

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