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人気のある微分積分 >

integral from 0 to 1 of sqrt(1+cos(x))

解

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

解

22 sin (\frac{1}{2})

+1

十進法表記

1.35602 \dots

解答ステップ

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

u置換積分法を適用する

= \int_{0}^{t a n (\frac{1}{2})} \frac{2 2}{( 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

= 22 \cdot \int_{0}^{t a n (\frac{1}{2})} \frac{1}{( 1 + u ^{2} ) 1 + u ^{2}} d u

簡素化

(1 + u^{2}) 1 + u^{2} : (1 + u^{2})^{\frac{3}{2}}

= 22 \cdot \int_{0}^{t a n (\frac{1}{2})} \frac{1}{( 1 + u ^{2} ) ^{\frac{3}{2}}} d u

三角関数による置換を適用する

= 22 \cdot \int_{0}^{\frac{1}{2}} cos (v) d v

共通積分を使用する:

\int cos (v) d v = sin (v)

= 22 [sin (v)]_{0}^{\frac{1}{2}}

限度を計算する:

sin (\frac{1}{2})

= 22 sin (\frac{1}{2})

グラフ

人気の例

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