integral from-3 to 3 of sqrt(x^2+y^2)

解

\int_{- 3}^{3} x^{2} + y^{2} d y

3 x^{2} + 9 + \frac{x ^{2} ( ln ( \frac{∣ 3 + x ^{2} + 9 ∣}{∣ x ∣} ) - ln ( \frac{∣ - 3 + x ^{2} + 9 ∣}{∣ x ∣} ) )}{2}

解答ステップ

\int_{- 3}^{3} x^{2} + y^{2} d y

三角関数による置換を適用する

= \int_{a r c t a n (- \frac{3}{x})}^{a r c t a n (\frac{3}{x})} x^{2} sec^{3} (u) d u

定数を除く:

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

= x^{2} \cdot \int_{a r c t a n (- \frac{3}{x})}^{a r c t a n (\frac{3}{x})} sec^{3} (u) d u

積分換算を適用する

= x^{2} [\frac{sec ^{2} ( u ) sin ( u )}{2}]_{a r c t a n (- \frac{3}{x})}^{a r c t a n (\frac{3}{x})} + \frac{1}{2} \cdot \int_{a r c t a n (- \frac{3}{x})}^{a r c t a n (\frac{3}{x})} sec (u) d u

\int_{a r c t a n (- \frac{3}{x})}^{a r c t a n (\frac{3}{x})} sec (u) d u = ln (\frac{3 + x ^{2} + 9}{∣ x ∣}) - ln (\frac{- 3 + x ^{2} + 9}{∣ x ∣})

= x^{2} [\frac{sec ^{2} ( u ) sin ( u )}{2}]_{a r c t a n (- \frac{3}{x})}^{a r c t a n (\frac{3}{x})} + \frac{1}{2} (ln (\frac{3 + x ^{2} + 9}{∣ x ∣}) - ln (\frac{- 3 + x ^{2} + 9}{∣ x ∣}))

簡素化

x^{2} [\frac{sec ^{2} ( u ) sin ( u )}{2}]_{a r c t a n (- \frac{3}{x})}^{a r c t a n (\frac{3}{x})} + \frac{1}{2} (ln (\frac{3 + x ^{2} + 9}{∣ x ∣}) - ln (\frac{- 3 + x ^{2} + 9}{∣ x ∣})) : x^{2} ([\frac{1}{2} sec (u) tan (u)]_{a r c t a n (- \frac{3}{x})}^{a r c t a n (\frac{3}{x})} + \frac{1}{2} (ln (\frac{3 + x ^{2} + 9}{∣ x ∣}) - ln (\frac{- 3 + x ^{2} + 9}{∣ x ∣})))

= x^{2} ([\frac{1}{2} sec (u) tan (u)]_{a r c t a n (- \frac{3}{x})}^{a r c t a n (\frac{3}{x})} + \frac{1}{2} (ln (\frac{3 + x ^{2} + 9}{∣ x ∣}) - ln (\frac{- 3 + x ^{2} + 9}{∣ x ∣})))

限度を計算する:

\frac{3 x ^{2} + 9}{x ^{2}}

= x^{2} (\frac{3 x ^{2} + 9}{x ^{2}} + \frac{1}{2} (ln (\frac{3 + x ^{2} + 9}{∣ x ∣}) - ln (\frac{- 3 + x ^{2} + 9}{∣ x ∣})))

簡素化

= 3 x^{2} + 9 + \frac{x ^{2} ( ln ( \frac{∣ 3 + x ^{2} + 9 ∣}{∣ x ∣} ) - ln ( \frac{∣ - 3 + x ^{2} + 9 ∣}{∣ x ∣} ) )}{2}