What is the integral of sin^2(z) ?

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

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integral of sin^2(z)

\int sin^{2} (z) d z

\frac{1}{2} (z - \frac{1}{2} sin (2 z)) + C

Solution steps

\int sin^{2} (z) d z

Rewrite using trig identities

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

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 z) d z

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 z - \int cos (2 z) d z)

\int 1 d z = z

\int cos (2 z) d z = \frac{1}{2} sin (2 z)

= \frac{1}{2} (z - \frac{1}{2} sin (2 z))

Add a constant to the solution

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

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