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

The integral of (sin^2(z))/2-sin(z) is 1/4 (z-1/2 sin(2z))+cos(z)+C

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integral of (sin^2(z))/2-sin(z)

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

\frac{1}{4} (z - \frac{1}{2} sin (2 z)) + cos (z) + C

Solution steps

\int \frac{sin ^{2} ( z )}{2} - sin (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

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

\int \frac{sin ^{2} ( z )}{2} d z = \frac{1}{4} (z - \frac{1}{2} sin (2 z))

\int sin (z) d z = - cos (z)

= \frac{1}{4} (z - \frac{1}{2} sin (2 z)) - (- cos (z))

Simplify

= \frac{1}{4} (z - \frac{1}{2} sin (2 z)) + cos (z)

Add a constant to the solution

= \frac{1}{4} (z - \frac{1}{2} sin (2 z)) + cos (z) + C

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What is the integral of (sin^2(z))/2-sin(z) ?
The integral of (sin^2(z))/2-sin(z) is 1/4 (z-1/2 sin(2z))+cos(z)+C