integral from 0 to a of sin(cx)sin(cx)

解

\int_{0}^{a} sin (c x) sin (c x) d x

\frac{1}{2} a - \frac{1}{4 c} sin (2 c a)

解答ステップ

\int_{0}^{a} sin (c x) sin (c x) d x

簡素化

sin (c x) sin (c x) : sin^{2} (c x)

= \int_{0}^{a} sin^{2} (c x) d x

三角関数の公式を使用して書き換える

= \int_{0}^{a} 1 - cos^{2} (c x) d x

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

= \int_{0}^{a} 1 d x - \int_{0}^{a} cos^{2} (c x) d x

\int_{0}^{a} 1 d x = a

\int_{0}^{a} cos^{2} (c x) d x = \frac{1}{2} (a + \frac{sin ( 2 c a )}{2 c})

= a - \frac{1}{2} (a + \frac{sin ( 2 c a )}{2 c})

簡素化

a - \frac{1}{2} (a + \frac{sin ( 2 c a )}{2 c}) : \frac{1}{2} a - \frac{1}{4 c} sin (2 c a)

= \frac{1}{2} a - \frac{1}{4 c} sin (2 c a)