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

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

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integral of sin^2((npix)/a)

\int sin^{2} (\frac{nπ x}{a}) d x

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

Solution steps

\int sin^{2} (\frac{nπ x}{a}) d x

Rewrite using trig identities

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

\int 1 d x = x

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

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

Add a constant to the solution

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

\int ln (x^{9}) d x

\frac{d}{d x} (\frac{8}{3})

x + 2, \frac{1}{x + 1}, 0, 2

\frac{d y}{d x} = \frac{( 2 x + 2 x y ^{2} )}{( y + 2 x ^{2} y )}

x^{2} + 9

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