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人気のある微分積分 >

integral of 5xsqrt(1-x^4)

解

\int 5 x 1 - x^{4} d x

解

\frac{5}{4} (arcsin (x^{2}) + \frac{1}{2} sin (2 arcsin (x^{2}))) + C

解答ステップ

\int 5 x 1 - x^{4} d x

定数を除く:

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

= 5 \cdot \int x 1 - x^{4} d x

簡素化

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

= 5 \cdot \int x 1 - (x^{2})^{2} d x

u置換積分法を適用する

= 5 \cdot \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

= 5 \cdot \frac{1}{2} \cdot \int 1 - u^{2} d u

三角関数による置換を適用する

= 5 \cdot \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

= 5 \cdot \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

= 5 \cdot \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

= 5 \cdot \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)

= 5 \cdot \frac{1}{2} \cdot \frac{1}{2} (v + \frac{1}{2} sin (2 v))

代用を戻す

= 5 \cdot \frac{1}{2} \cdot \frac{1}{2} (arcsin (x^{2}) + \frac{1}{2} sin (2 arcsin (x^{2})))

簡素化

5 \cdot \frac{1}{2} \cdot \frac{1}{2} (arcsin (x^{2}) + \frac{1}{2} sin (2 arcsin (x^{2}))) : \frac{5}{4} (arcsin (x^{2}) + \frac{1}{2} sin (2 arcsin (x^{2})))

= \frac{5}{4} (arcsin (x^{2}) + \frac{1}{2} sin (2 arcsin (x^{2})))

定数を解答に追加する

= \frac{5}{4} (arcsin (x^{2}) + \frac{1}{2} sin (2 arcsin (x^{2}))) + C

グラフ

人気の例

derivative x(x^2+4)^5

d er i v a t i v e x (x^{2} + 4)^{5}

derivative f(x)=5x+2x^2+7e^x

d er i v a t i v e f (x) = 5 x + 2 x^{2} + 7 e^{x}

面積 y=-x^2+1,y=0

a re a y = - x^{2} + 1, y = 0

integral from 0 to pi of 1/(sqrt(pi-x))

\int_{0}^{π} \frac{1}{π - x} d x

limit as x approaches-5/2 of-x+2

x \to - \frac{5}{2} lim (- x + 2)