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人気のある微分積分 >

integral from 0 to 1 of sqrt(2-2cos(x))

解

\int_{0}^{1} 2 - 2 cos (x) d x

解

2 (- \frac{2}{sec ( \frac{1}{2} )} + 2)

+1

十進法表記

0.48966 \dots

解答ステップ

\int_{0}^{1} 2 - 2 cos (x) d x

u置換積分法を適用する

= \int_{0}^{t a n (\frac{1}{2})} \frac{4 u}{( 1 + u ^{2} ) 1 + u ^{2}} d u

定数を除く:

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

= 4 \cdot \int_{0}^{t a n (\frac{1}{2})} \frac{u}{( 1 + u ^{2} ) 1 + u ^{2}} d u

u置換積分法を適用する

= 4 \cdot \int_{1}^{1 + t a n^{2} (\frac{1}{2})} \frac{1}{2 v v} d v

定数を除く:

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

= 4 \cdot \frac{1}{2} \cdot \int_{1}^{1 + t a n^{2} (\frac{1}{2})} \frac{1}{v v} d v

簡素化

v v : v^{\frac{3}{2}}

= 4 \cdot \frac{1}{2} \cdot \int_{1}^{1 + t a n^{2} (\frac{1}{2})} \frac{1}{v ^{\frac{3}{2}}} d v

べき乗則を適用する

= 4 \cdot \frac{1}{2} [- \frac{2}{v}]_{1}^{1 + t a n^{2} (\frac{1}{2})}

簡素化

4 \cdot \frac{1}{2} [- \frac{2}{v}]_{1}^{1 + t a n^{2} (\frac{1}{2})} : 2 [- \frac{2}{v}]_{1}^{s e c^{2} (\frac{1}{2})}

= 2 [- \frac{2}{v}]_{1}^{s e c^{2} (\frac{1}{2})}

限度を計算する:

- \frac{2}{sec ( \frac{1}{2} )} + 2

= 2 (- \frac{2}{sec ( \frac{1}{2} )} + 2)

グラフ

人気の例

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