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人気のある微分積分 >

integral of sqrt(4-x^2-y^2)

解

\int 4 - x^{2} - y^{2} d x

解

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

解答ステップ

\int 4 - x^{2} - y^{2} d x

三角関数による置換を適用する

= \int cos^{2} (u) (4 - y^{2}) d u

定数を除く:

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

= (4 - y^{2}) \cdot \int cos^{2} (u) d u

\int cos^{2} (u) d u = \frac{1}{2} (u + \frac{1}{2} sin (2 u))

= (4 - y^{2}) \frac{1}{2} (u + \frac{1}{2} sin (2 u))

代用を戻す

u = arcsin (\frac{1}{4 - y ^{2}} x)

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

簡素化

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

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

定数を解答に追加する

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

グラフ

人気の例

integral of x^2ln(1+x)

\int x^{2} ln (1 + x) d x

integral of 4xcos(4x^2)

\int 4 x cos (4 x^{2}) d x

(dy)/(dx)=(y^3)/(3x-2),y(1)=-1

\frac{d y}{d x} = \frac{y ^{3}}{3 x - 2}, y (1) = - 1

integral from-infinity to 5 of 1/(x^2+1)

\int_{- \infty}^{5} \frac{1}{x ^{2} + 1} d x

integral of 1+cos(θ)

\int 1 + cos (θ) d θ