integral from 1 to 4 of 3/2 sin(2-x/2)

解

\int_{1}^{4} \frac{3}{2} sin (2 - \frac{x}{2}) d x

3 cos^{2} (2) - 3 cos (2) cos (\frac{1}{2}) + 3 sin^{2} (2) - 3 sin (2) sin (\frac{1}{2})

十進法表記

2.78778 \dots

解答ステップ

\int_{1}^{4} \frac{3}{2} sin (2 - \frac{x}{2}) d x

定数を除く:

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

= \frac{3}{2} \cdot \int_{1}^{4} sin (2 - \frac{x}{2}) d x

三角関数の公式を使用して書き換える

= \frac{3}{2} \cdot \int_{1}^{4} - cos (2) sin (\frac{x}{2}) + cos (\frac{x}{2}) sin (2) 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

= \frac{3}{2} (- \int_{1}^{4} cos (2) sin (\frac{x}{2}) d x + \int_{1}^{4} cos (\frac{x}{2}) sin (2) d x)

\int_{1}^{4} cos (2) sin (\frac{x}{2}) d x = - 2 cos^{2} (2) + 2 cos (2) cos (\frac{1}{2})

\int_{1}^{4} cos (\frac{x}{2}) sin (2) d x = 2 sin^{2} (2) - 2 sin (2) sin (\frac{1}{2})

= \frac{3}{2} (- (- 2 cos^{2} (2) + 2 cos (2) cos (\frac{1}{2})) + 2 sin^{2} (2) - 2 sin (2) sin (\frac{1}{2}))

簡素化

\frac{3}{2} (- (- 2 cos^{2} (2) + 2 cos (2) cos (\frac{1}{2})) + 2 sin^{2} (2) - 2 sin (2) sin (\frac{1}{2})) : 3 cos^{2} (2) - 3 cos (2) cos (\frac{1}{2}) + 3 sin^{2} (2) - 3 sin (2) sin (\frac{1}{2})

= 3 cos^{2} (2) - 3 cos (2) cos (\frac{1}{2}) + 3 sin^{2} (2) - 3 sin (2) sin (\frac{1}{2})