integral from 1 to 2 of (1+cos(x))^2

解

\int_{1}^{2} (1 + cos (x))^{2} d x

- 2 sin (1) + \frac{3}{2} + \frac{sin ( 4 ) + 7 sin ( 2 )}{4}

十進法表記

1.21912 \dots

解答ステップ

\int_{1}^{2} (1 + cos (x))^{2} d x

拡張

(1 + cos (x))^{2} : 1 + 2 cos (x) + cos^{2} (x)

= \int_{1}^{2} 1 + 2 cos (x) + cos^{2} (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_{1}^{2} 1 d x + \int_{1}^{2} 2 cos (x) d x + \int_{1}^{2} cos^{2} (x) d x

\int_{1}^{2} 1 d x = 1

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

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

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

簡素化

1 + 2 (sin (2) - sin (1)) + \frac{1}{2} (1 + \frac{1}{2} (sin (4) - sin (2))) : - 2 sin (1) + \frac{3}{2} + \frac{sin ( 4 ) + 7 sin ( 2 )}{4}

= - 2 sin (1) + \frac{3}{2} + \frac{sin ( 4 ) + 7 sin ( 2 )}{4}