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

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

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integral of sin^2(npix)

\int sin^{2} (nπ x) d x

\frac{x}{2} - \frac{1}{4 πn} sin (2 πn x) + C

Solution steps

\int sin^{2} (nπ x) d x

Rewrite using trig identities

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

Apply u-substitution

: \int \frac{1 - cos ( u )}{2 πn} d u

= \frac{1}{2} \cdot \int \frac{1 - cos ( u )}{2 πn} d u

Take the constant out:

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

= \frac{1}{2} \cdot \frac{1}{2 πn} \cdot \int 1 - cos (u) d u

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} \cdot \frac{1}{2 πn} (\int 1 d u - \int cos (u) d u)

\int 1 d u = u

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

= \frac{1}{2} \cdot \frac{1}{2 πn} (u - sin (u))

Substitute back

u = 2 nπ x

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

Simplify

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

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

Add a constant to the solution

= \frac{x}{2} - \frac{1}{4 πn} sin (2 πn x) + C

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