What is the integral of sin^2((pix)/L) ?

The integral of sin^2((pix)/L) is x/2-L/(4pi)sin((2pix)/L)+C

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integral of sin^2((pix)/L)

\int sin^{2} (\frac{π x}{L}) d x

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

Solution steps

\int sin^{2} (\frac{π x}{L}) d x

Apply u-substitution

: \int \frac{L sin ^{2} ( u )}{π} d u

= \int \frac{L sin ^{2} ( u )}{π} d u

Take the constant out:

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

= \frac{L}{π} \cdot \int sin^{2} (u) d u

Rewrite using trig identities

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

Substitute back

u = \frac{x π}{L}

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

Simplify

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

= \frac{x}{2} - \frac{L}{4 π} sin (\frac{2 π x}{L})

Add a constant to the solution

= \frac{x}{2} - \frac{L}{4 π} sin (\frac{2 π x}{L}) + C

t^{2} y^{^{''}} - 3 t y^{^{'}} + 4 y = 0

\frac{\partial}{\partial y} ((x + y)^{5} + (x - y)^{5})

f (x) = x^{2} + 5, (- 5, 30)

x y^{^{''}} - 2 y^{^{'}} + 4 = 0

(- 3.7) (- 4.12)

What is the integral of sin^2((pix)/L) ?
The integral of sin^2((pix)/L) is x/2-L/(4pi)sin((2pix)/L)+C