ja

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

人気のある微分積分 >

integral from pi/6 to pi/3 of 5csc^3(x)

解

\int_{\frac{π}{6}}^{\frac{π}{3}} 5 csc^{3} (x) d x

解

\frac{5 ( - 3 ln ( 3 ( 7 - 4 3 ) ) + 12 3 - 4 )}{12}

+1

十進法表記

8.91271 \dots

解答ステップ

\int_{\frac{π}{6}}^{\frac{π}{3}} 5 csc^{3} (x) d x

定数を除く:

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

= 5 \cdot \int_{\frac{π}{6}}^{\frac{π}{3}} csc^{3} (x) d x

u置換積分法を適用する

= 5 \cdot \int_{2 - 3}^{\frac{3}{3}} \frac{( 1 + u ^{2} ) ^{2}}{4 u ^{3}} 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}{4} \cdot \int_{2 - 3}^{\frac{3}{3}} \frac{( 1 + u ^{2} ) ^{2}}{u ^{3}} d u

拡張

\frac{( 1 + u ^{2} ) ^{2}}{u ^{3}} : \frac{1}{u ^{3}} + \frac{2}{u} + u

= 5 \cdot \frac{1}{4} \cdot \int_{2 - 3}^{\frac{3}{3}} \frac{1}{u ^{3}} + \frac{2}{u} + u d u

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}{4} (\int_{2 - 3}^{\frac{3}{3}} \frac{1}{u ^{3}} d u + \int_{2 - 3}^{\frac{3}{3}} \frac{2}{u} d u + \int_{2 - 3}^{\frac{3}{3}} u d u)

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

\int_{2 - 3}^{\frac{3}{3}} \frac{2}{u} d u = - ln (3 (7 - 43))

\int_{2 - 3}^{\frac{3}{3}} u d u = \frac{1}{6} - \frac{7 - 4 3}{2}

= 5 \cdot \frac{1}{4} (2 (1 + 3) - ln (3 (7 - 43)) + \frac{1}{6} - \frac{7 - 4 3}{2})

簡素化

5 \cdot \frac{1}{4} (2 (1 + 3) - ln (3 (7 - 43)) + \frac{1}{6} - \frac{7 - 4 3}{2}) : \frac{5 ( - 3 ln ( 3 ( 7 - 4 3 ) ) + 12 3 - 4 )}{12}

= \frac{5 ( - 3 ln ( 3 ( 7 - 4 3 ) ) + 12 3 - 4 )}{12}

グラフ

人気の例

integral from 0 to x of cosh^2(t)

\int_{0}^{x} cosh^{2} (t) d t

integral from 0 to x of 1/(1+t+t^2)

\int_{0}^{x} \frac{1}{1 + t + t ^{2}} d t

integral from 0 to 2 of 1/((x-2)^{3/5)}

\int_{0}^{2} \frac{1}{( x - 2 ) ^{\frac{3}{5}}} d x

integral from 0 to 2 of 3^{4x}

\int_{0}^{2} 3^{4 x} d x

integral from sqrt(x) to 0 of cos(t^2)

\int_{x}^{0} cos (t^{2}) d t