What is the integral of sin^2((npi)/a x) ?

The integral of sin^2((npi)/a x) is x/2-a/(4pin)sin(2*(pin)/a x)+C

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integral of sin^2((npi)/a x)

\int sin^{2} (\frac{nπ}{a} x) d x

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

Solution steps

\int sin^{2} (\frac{nπ}{a} x) d x

Apply u-substitution

= \int \frac{a sin ^{2} ( u )}{πn} d u

Take the constant out:

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

= \frac{a}{πn} \cdot \int sin^{2} (u) d u

Rewrite using trig identities

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

Take the constant out:

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

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

\int 1 d u = u

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

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

Substitute back

u = \frac{nπ}{a} x

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

Simplify

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

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

Add a constant to the solution

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

What is the integral of sin^2((npi)/a x) ?
The integral of sin^2((npi)/a x) is x/2-a/(4pin)sin(2*(pin)/a x)+C