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人気のある微分積分 >

integral of x^2sqrt(3+2x-x^2)

解

\int x^{2} 3 + 2 x - x^{2} d x

解

4 (arcsin (\frac{1}{2} (x - 1)) + \frac{1}{16} (x - 1)^{3} - x^{2} + 2 x + 3 - \frac{1}{6} (- x^{2} + 2 x + 3)^{\frac{3}{2}}) + C

解答ステップ

\int x^{2} 3 + 2 x - x^{2} d x

平方完成する

3 + 2 x - x^{2} : - (x - 1)^{2} + 4

= \int x^{2} - (x - 1)^{2} + 4 d x

u置換積分法を適用する

= \int (u + 1)^{2} - u^{2} + 4 d u

三角関数による置換を適用する

= \int 4 cos^{2} (v) (2 sin (v) + 1)^{2} d v

定数を除く:

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

= 4 \cdot \int cos^{2} (v) (2 sin (v) + 1)^{2} d v

三角関数の公式を使用して書き換える

= 4 \cdot \int (1 - sin^{2} (v)) (2 sin (v) + 1)^{2} d v

拡張

(1 - sin^{2} (v)) (2 sin (v) + 1)^{2} : 3 sin^{2} (v) + 4 sin (v) - 4 sin^{4} (v) - 4 sin^{3} (v) + 1

= 4 \cdot \int 3 sin^{2} (v) + 4 sin (v) - 4 sin^{4} (v) - 4 sin^{3} (v) + 1 d v

総和規則を適用する:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= 4 (\int 3 sin^{2} (v) d v + \int 4 sin (v) d v - \int 4 sin^{4} (v) d v - \int 4 sin^{3} (v) d v + \int 1 d v)

\int 3 sin^{2} (v) d v = \frac{3}{2} (v - \frac{1}{2} sin (2 v))

\int 4 sin (v) d v = - 4 cos (v)

\int 4 sin^{4} (v) d v = - sin^{3} (v) cos (v) + \frac{3}{2} v - \frac{3}{4} sin (2 v)

\int 4 sin^{3} (v) d v = 4 (- cos (v) + \frac{cos ^{3} ( v )}{3})

\int 1 d v = v

= 4 (\frac{3}{2} (v - \frac{1}{2} sin (2 v)) - 4 cos (v) - (- sin^{3} (v) cos (v) + \frac{3}{2} v - \frac{3}{4} sin (2 v)) - 4 (- cos (v) + \frac{cos ^{3} ( v )}{3}) + v)

代用を戻す

= 4 (\frac{3}{2} (arcsin (\frac{1}{2} (x - 1)) - \frac{1}{2} sin (2 arcsin (\frac{1}{2} (x - 1)))) - 4 cos (arcsin (\frac{1}{2} (x - 1))) - (- sin^{3} (arcsin (\frac{1}{2} (x - 1))) cos (arcsin (\frac{1}{2} (x - 1))) + \frac{3}{2} arcsin (\frac{1}{2} (x - 1)) - \frac{3}{4} sin (2 arcsin (\frac{1}{2} (x - 1)))) - 4 (- cos (arcsin (\frac{1}{2} (x - 1))) + \frac{cos ^{3} ( arcsin ( \frac{1}{2} ( x - 1 ) ) )}{3}) + arcsin (\frac{1}{2} (x - 1)))

簡素化

4 (\frac{3}{2} (arcsin (\frac{1}{2} (x - 1)) - \frac{1}{2} sin (2 arcsin (\frac{1}{2} (x - 1)))) - 4 cos (arcsin (\frac{1}{2} (x - 1))) - (- sin^{3} (arcsin (\frac{1}{2} (x - 1))) cos (arcsin (\frac{1}{2} (x - 1))) + \frac{3}{2} arcsin (\frac{1}{2} (x - 1)) - \frac{3}{4} sin (2 arcsin (\frac{1}{2} (x - 1)))) - 4 (- cos (arcsin (\frac{1}{2} (x - 1))) + \frac{cos ^{3} ( arcsin ( \frac{1}{2} ( x - 1 ) ) )}{3}) + arcsin (\frac{1}{2} (x - 1))) : 4 (arcsin (\frac{1}{2} (x - 1)) + \frac{1}{16} (x - 1)^{3} - x^{2} + 2 x + 3 - \frac{1}{6} (- x^{2} + 2 x + 3)^{\frac{3}{2}})

= 4 (arcsin (\frac{1}{2} (x - 1)) + \frac{1}{16} (x - 1)^{3} - x^{2} + 2 x + 3 - \frac{1}{6} (- x^{2} + 2 x + 3)^{\frac{3}{2}})

定数を解答に追加する

= 4 (arcsin (\frac{1}{2} (x - 1)) + \frac{1}{16} (x - 1)^{3} - x^{2} + 2 x + 3 - \frac{1}{6} (- x^{2} + 2 x + 3)^{\frac{3}{2}}) + C

グラフ

人気の例

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\frac{d}{d x} (4 e^{\frac{x}{2}} - 3 e^{- x})

integral from 0 to pi/2 of xsin(8x)

\int_{0}^{\frac{π}{2}} x sin (8 x) d x

tangent f(x)=6x-x^2

t an g e n t f (x) = 6 x - x^{2}

8-x^2,x^2,x=-3,x=3

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y^'=5y+e^{-2x}y^{-2}

y^{^{'}} = 5 y + e^{- 2 x} y^{- 2}