What is the integral of 1/4 sin^2(x) ?

The integral of 1/4 sin^2(x) is 1/8 (x-1/2 sin(2x))+C

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integral of 1/4 sin^2(x)

\int \frac{1}{4} sin^{2} (x) d x

\frac{1}{8} (x - \frac{1}{2} sin (2 x)) + C

Solution steps

\int \frac{1}{4} sin^{2} (x) d x

Take the constant out:

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

= \frac{1}{4} \cdot \int sin^{2} (x) d x

Rewrite using trig identities

= \frac{1}{4} \cdot \int \frac{1 - cos ( 2 x )}{2} d x

Take the constant out:

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

= \frac{1}{4} \cdot \frac{1}{2} \cdot \int 1 - cos (2 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}{4} \cdot \frac{1}{2} (\int 1 d x - \int cos (2 x) d x)

\int 1 d x = x

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

= \frac{1}{4} \cdot \frac{1}{2} (x - \frac{1}{2} sin (2 x))

Simplify

\frac{1}{4} \cdot \frac{1}{2} (x - \frac{1}{2} sin (2 x)) : \frac{1}{8} (x - \frac{1}{2} sin (2 x))

= \frac{1}{8} (x - \frac{1}{2} sin (2 x))

Add a constant to the solution

= \frac{1}{8} (x - \frac{1}{2} sin (2 x)) + C

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What is the integral of 1/4 sin^2(x) ?
The integral of 1/4 sin^2(x) is 1/8 (x-1/2 sin(2x))+C