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人気のある微分積分 >

integral of (x^5sqrt(1-x^2))

解

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

解

- \frac{1}{3} (1 - x^{2})^{\frac{3}{2}} + \frac{2}{5} (1 - x^{2})^{\frac{5}{2}} - \frac{1}{7} (1 - x^{2})^{\frac{7}{2}} + C

解答ステップ

\int x^{5} 1 - x^{2} d x

三角関数による置換を適用する

= \int sin^{5} (u) cos^{2} (u) d u

三角関数の公式を使用して書き換える

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

u置換積分法を適用する

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

拡張

- v^{2} (1 - v^{2})^{2} : - v^{2} + 2 v^{4} - v^{6}

= \int - v^{2} + 2 v^{4} - v^{6} 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

= - \int v^{2} d v + \int 2 v^{4} d v - \int v^{6} d v

\int v^{2} d v = \frac{v ^{3}}{3}

\int 2 v^{4} d v = \frac{2 v ^{5}}{5}

\int v^{6} d v = \frac{v ^{7}}{7}

= - \frac{v ^{3}}{3} + \frac{2 v ^{5}}{5} - \frac{v ^{7}}{7}

代用を戻す

= - \frac{cos ^{3} ( arcsin ( x ) )}{3} + \frac{2 cos ^{5} ( arcsin ( x ) )}{5} - \frac{cos ^{7} ( arcsin ( x ) )}{7}

簡素化

- \frac{cos ^{3} ( arcsin ( x ) )}{3} + \frac{2 cos ^{5} ( arcsin ( x ) )}{5} - \frac{cos ^{7} ( arcsin ( x ) )}{7} : - \frac{1}{3} (1 - x^{2})^{\frac{3}{2}} + \frac{2}{5} (1 - x^{2})^{\frac{5}{2}} - \frac{1}{7} (1 - x^{2})^{\frac{7}{2}}

= - \frac{1}{3} (1 - x^{2})^{\frac{3}{2}} + \frac{2}{5} (1 - x^{2})^{\frac{5}{2}} - \frac{1}{7} (1 - x^{2})^{\frac{7}{2}}

定数を解答に追加する

= - \frac{1}{3} (1 - x^{2})^{\frac{3}{2}} + \frac{2}{5} (1 - x^{2})^{\frac{5}{2}} - \frac{1}{7} (1 - x^{2})^{\frac{7}{2}} + C

グラフ

人気の例

integral of (2t-5)/(4t^2+1)

\int \frac{2 t - 5}{4 t ^{2} + 1} d t

integral of tan(7z)

\int tan (7 z) d z

integral of 1/2 xsec(x)tan(x)

\int \frac{1}{2} x sec (x) tan (x) d x

integral of (5x)/(sqrt(x^2-1))

\int \frac{5 x}{x ^{2} - 1} d x

integral of (2x-1)/((2x^2-2x+4)^2)

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