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人気のある微分積分 >

integral from pi/2 to 0 of-sin^2(t)

解

\int_{\frac{π}{2}}^{0} - sin^{2} (t) d t

解

\frac{π}{4}

+1

十進法表記

0.78539 \dots

解答ステップ

\int_{\frac{π}{2}}^{0} - sin^{2} (t) d t

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

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

定数を除く:

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

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

三角関数の公式を使用して書き換える

= - (- \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}

グラフ

人気の例

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