integral of sin^2(x)/2 — step-by-step solution

1) Let's integrate:

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

We reduce the square using the identity

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

ⅰ

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

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

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

ⅰ

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

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

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

For the math expression:

\frac{\cos (2 x)}{4}

, since (sin(a))' = a'*cos(a), we try differentiating

\sin (2 x)

2) Let's perform the differentiation:

\frac{d}{d x} (\sin (2 x))

Apply the differentiation rule:

\frac{d}{d x} (\sin (f)) = \frac{d}{d x} (f) \cdot \cos (f)

with f =

2 x

ⓘ The derivative of sin(x) is cos(x):

\frac{d}{d x} (\sin (2 x))

\frac{d}{d x} (2 x) \cdot \cos (2 x)

\frac{d}{d x} (2 x) \cdot \cos (2 x)

ⓘ A constant factor stays in front of the derivative, (k*u)' = k*u':

\frac{d}{d x} (2 x)

2

2 \cdot \cos (2 x)

You observe that the computed derivative and the original math expression are the same, except for a coefficient. So the integral is:

ⓘ The derivative of

\frac{1}{4} \cdot x

\frac{1}{4}

, so

\frac{1}{4} \cdot x

is an antiderivative of

\frac{1}{4}

\int \frac{1}{4} d x

\frac{1}{4} \cdot x

■The derivative of

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

- \frac{\cos (2 x)}{4}

, so

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

is an antiderivative of

- \frac{\cos (2 x)}{4}

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

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

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

ⓘ Multiplying a fraction by a term multiplies its numerator:

\frac{1}{4} \cdot x

\frac{1 \cdot x}{4}

\frac{x}{4}

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

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

\frac{\sin (2 x)}{8}

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

▶Indefinite integrals are defined up to an additive constant C, so this yields:

\frac{x}{4} - \frac{\sin (2 x)}{8} + C