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人気のある微分積分 >

integral of (1-4x^2)^{3/2}

解

\int (1 - 4 x^{2})^{\frac{3}{2}} d x

解

x (1 - 4 x^{2})^{\frac{3}{2}} + \frac{3}{16} arcsin (2 x) - \frac{3}{64} sin (4 arcsin (2 x)) + C

解答ステップ

\int (1 - 4 x^{2})^{\frac{3}{2}} d x

置換積分法を適用する

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

部分積分を適用する

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

\int - 3 u^{2} (1 - u^{2})^{\frac{1}{2}} d u = - \frac{3}{8} (arcsin (u) - \frac{1}{4} sin (4 arcsin (u)))

= \frac{1}{2} (u (1 - u^{2})^{\frac{3}{2}} - (- \frac{3}{8} (arcsin (u) - \frac{1}{4} sin (4 arcsin (u)))))

代用を戻す

u = 2 x

= \frac{1}{2} (2 x (1 - (2 x)^{2})^{\frac{3}{2}} - (- \frac{3}{8} (arcsin (2 x) - \frac{1}{4} sin (4 arcsin (2 x)))))

簡素化

\frac{1}{2} (2 x (1 - (2 x)^{2})^{\frac{3}{2}} - (- \frac{3}{8} (arcsin (2 x) - \frac{1}{4} sin (4 arcsin (2 x))))) : x (1 - 4 x^{2})^{\frac{3}{2}} + \frac{3}{16} arcsin (2 x) - \frac{3}{64} sin (4 arcsin (2 x))

= x (1 - 4 x^{2})^{\frac{3}{2}} + \frac{3}{16} arcsin (2 x) - \frac{3}{64} sin (4 arcsin (2 x))

定数を解答に追加する

= x (1 - 4 x^{2})^{\frac{3}{2}} + \frac{3}{16} arcsin (2 x) - \frac{3}{64} sin (4 arcsin (2 x)) + C

グラフ

人気の例

integral of (5x^2-2x+64)/(x^3-16x)

\int \frac{5 x ^{2} - 2 x + 64}{x ^{3} - 16 x} d x

integral of (e^{2x}+1)^{10}e^{2x}

\int (e^{2 x} + 1)^{10} e^{2 x} d x

integral of sqrt(1+((y-1)^2)/(4y))

\int 1 + \frac{( y - 1 ) ^{2}}{4 y} d y

integral of 6x(3x^2-1)

\int 6 x (3 x^{2} - 1) d x

積分記号 1/(x^2+9)

in t e g r a l \frac{1}{x ^{2} + 9}