What is the integral from 0 to L of sin^2((npi)/L x) ?

The integral from 0 to L of sin^2((npi)/L x) is L/2-L/(4pin)sin(2pin)

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integral from 0 to L of sin^2((npi)/L x)

\int_{0}^{L} sin^{2} (\frac{nπ}{L} x) d x

\frac{L}{2} - \frac{L}{4 πn} sin (2 πn)

Solution steps

\int_{0}^{L} sin^{2} (\frac{nπ}{L} x) d x

Apply u-substitution

: \int_{0}^{πn} \frac{L sin ^{2} ( u )}{πn} d u

= \int_{0}^{πn} \frac{L 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{L}{πn} \cdot \int_{0}^{πn} sin^{2} (u) d u

Rewrite using trig identities

= \frac{L}{πn} \cdot \int_{0}^{πn} \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}{πn} \cdot \frac{1}{2} \cdot \int_{0}^{πn} 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}{πn} \cdot \frac{1}{2} (\int_{0}^{πn} 1 d u - \int_{0}^{πn} cos (2 u) d u)

\int_{0}^{πn} 1 d u = πn

\int_{0}^{πn} cos (2 u) d u = \frac{1}{2} sin (2 πn)

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

Simplify

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

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

\int_{π}^{3 π} x^{2} sin (\frac{x}{2}) d x

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\int_{- 2}^{1} \frac{1}{( 1 + 5 x ) ^{3}} d x

\int_{1}^{4} (x - 1)^{2} d x

What is the integral from 0 to L of sin^2((npi)/L x) ?
The integral from 0 to L of sin^2((npi)/L x) is L/2-L/(4pin)sin(2pin)