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人気のある微分積分 >

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

解

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

解

\frac{π}{3 3}

+1

十進法表記

0.60459 \dots

解答ステップ

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

u置換積分法を適用する

= \int_{0}^{1} \frac{2}{u ^{2} + 3} d u

定数を除く:

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

= 2 \cdot \int_{0}^{1} \frac{1}{u ^{2} + 3} d u

置換積分法を適用する

= 2 \cdot \int_{0}^{\frac{1}{3}} \frac{1}{3 ( v ^{2} + 1 )} d v

定数を除く:

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

= 2 \cdot \frac{1}{3} \cdot \int_{0}^{\frac{1}{3}} \frac{1}{v ^{2} + 1} d v

共通積分を使用する:

\int \frac{1}{v ^{2} + 1} d v = arctan (v)

= 2 \cdot \frac{1}{3} [arctan (v)]_{0}^{\frac{1}{3}}

簡素化

2 \cdot \frac{1}{3} [arctan (v)]_{0}^{\frac{1}{3}} : \frac{2}{3} [arctan (v)]_{0}^{\frac{1}{3}}

= \frac{2}{3} [arctan (v)]_{0}^{\frac{1}{3}}

限度を計算する:

\frac{π}{6}

= \frac{2}{3} \cdot \frac{π}{6}

簡素化

= \frac{π}{3 3}

グラフ

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