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人気のある微分積分 >

integral from 0 to pi/2 of 1/(3+2cos(x))

解

\int_{0}^{\frac{π}{2}} \frac{1}{3 + 2 cos ( x )} d x

解

\frac{2 arctan ( \frac{1}{5} )}{5}

+1

十進法表記

0.37613 \dots

解答ステップ

\int_{0}^{\frac{π}{2}} \frac{1}{3 + 2 cos ( x )} d x

u置換積分法を適用する

= \int_{0}^{1} \frac{2}{u ^{2} + 5} 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} + 5} d u

置換積分法を適用する

= 2 \cdot \int_{0}^{\frac{1}{5}} \frac{1}{5 ( 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}{5} \cdot \int_{0}^{\frac{1}{5}} \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}{5} [arctan (v)]_{0}^{\frac{1}{5}}

簡素化

2 \cdot \frac{1}{5} [arctan (v)]_{0}^{\frac{1}{5}} : \frac{2}{5} [arctan (v)]_{0}^{\frac{1}{5}}

= \frac{2}{5} [arctan (v)]_{0}^{\frac{1}{5}}

限度を計算する:

arctan (\frac{1}{5})

= \frac{2}{5} arctan (\frac{1}{5})

簡素化

= \frac{2 arctan ( \frac{1}{5} )}{5}

グラフ

人気の例

integral from 0 to pi/4 of arctan(x)

\int_{0}^{\frac{π}{4}} arctan (x) d x

integral from 3 to 6 of x^2-6x+9

\int_{3}^{6} x^{2} - 6 x + 9 d x

integral from 0 to 2x of e^{t^2}

\int_{0}^{2 x} e^{t^{2}} d t

integral from pi/6 to pi/4 of sin(x)

\int_{\frac{π}{6}}^{\frac{π}{4}} sin (x) d x

integral from 1 to 2 of xe^{-x}

\int_{1}^{2} x e^{- x} d x