integral from 0 to pi/4 of 1-4sin^2(x)

解

\int_{0}^{\frac{π}{4}} 1 - 4 sin^{2} (x) d x

- \frac{π}{4} + 1

十進法表記

0.21460 \dots

解答ステップ

\int_{0}^{\frac{π}{4}} 1 - 4 sin^{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_{0}^{\frac{π}{4}} 1 d x - \int_{0}^{\frac{π}{4}} 4 sin^{2} (x) d x

\int_{0}^{\frac{π}{4}} 1 d x = \frac{π}{4}

\int_{0}^{\frac{π}{4}} 4 sin^{2} (x) d x = \frac{π}{2} - 1

= \frac{π}{4} - (\frac{π}{2} - 1)

簡素化

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

= - \frac{π}{4} + 1