integral from pi/2 to 0 of (1+cos(2t))/2

解

\int_{\frac{π}{2}}^{0} \frac{1 + cos ( 2 t )}{2} d t

- \frac{π}{4}

十進法表記

- 0.78539 \dots

解答ステップ

\int_{\frac{π}{2}}^{0} \frac{1 + cos ( 2 t )}{2} d t

\int_{a}^{b} f (x) d x = - \int_{b}^{a} f (x) d x, a < b

= - \int_{0}^{\frac{π}{2}} \frac{1 + cos ( 2 t )}{2} d t

定数を除く:

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

= - \frac{1}{2} \cdot \int_{0}^{\frac{π}{2}} 1 + cos (2 t) d t

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{1}{2} (\int_{0}^{\frac{π}{2}} 1 d t + \int_{0}^{\frac{π}{2}} cos (2 t) d t)

\int_{0}^{\frac{π}{2}} 1 d t = \frac{π}{2}

\int_{0}^{\frac{π}{2}} cos (2 t) d t = 0

= - \frac{1}{2} (\frac{π}{2} + 0)

簡素化

- \frac{1}{2} (\frac{π}{2} + 0) : - \frac{π}{4}

= - \frac{π}{4}