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人気のある微分積分 >

integral of (x*(sqrt(1-x^4)))

解

\int (x \cdot (1 - x^{4})) d x

解

\frac{1}{4} (arcsin (x^{2}) + \frac{1}{2} sin (2 arcsin (x^{2}))) + C

解答ステップ

\int x 1 - x^{4} d x

簡素化

x^{4} : (x^{2})^{2}

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

u置換積分法を適用する

= \int \frac{1 - u ^{2}}{2} d u

定数を除く:

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

= \frac{1}{2} \cdot \int 1 - u^{2} d u

三角関数による置換を適用する

= \frac{1}{2} \cdot \int cos^{2} (v) d v

三角関数の公式を使用して書き換える

= \frac{1}{2} \cdot \int \frac{1 + cos ( 2 v )}{2} d v

定数を除く:

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

= \frac{1}{2} \cdot \frac{1}{2} \cdot \int 1 + cos (2 v) 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

= \frac{1}{2} \cdot \frac{1}{2} (\int 1 d v + \int cos (2 v) d v)

\int 1 d v = v

\int cos (2 v) d v = \frac{1}{2} sin (2 v)

= \frac{1}{2} \cdot \frac{1}{2} (v + \frac{1}{2} sin (2 v))

代用を戻す

= \frac{1}{2} \cdot \frac{1}{2} (arcsin (x^{2}) + \frac{1}{2} sin (2 arcsin (x^{2})))

簡素化

\frac{1}{2} \cdot \frac{1}{2} (arcsin (x^{2}) + \frac{1}{2} sin (2 arcsin (x^{2}))) : \frac{1}{4} (arcsin (x^{2}) + \frac{1}{2} sin (2 arcsin (x^{2})))

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

定数を解答に追加する

= \frac{1}{4} (arcsin (x^{2}) + \frac{1}{2} sin (2 arcsin (x^{2}))) + C

グラフ

人気の例

(dy)/(dx)=y^2x^3

\frac{d y}{d x} = y^{2} x^{3}

integral of 2/(2+e^x)

\int \frac{2}{2 + e ^{x}} d x

limit as x approaches 3 of 7-3x

x \to 3 lim (7 - 3 x)

(\partial)/(\partial x)(-3x^2-4x)

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limit as x approaches 0+of 1-ln(x)

x \to 0 + lim (1 - ln (x))