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

解

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

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

解答ステップ

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

u置換積分法を適用する

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

定数を除く:

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

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

三角関数の公式を使用して書き換える

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

定数を除く:

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

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

総和規則を適用する:

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

簡素化

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

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