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人気のある微分積分 >

integral from (5pi)/6 to pi of (cos^4(x))/(sqrt(1-sin(x)))

解

\int_{\frac{5 π}{6}}^{π} \frac{cos ^{4} ( x )}{1 - sin ( x )} d x

解

- \frac{18}{35} - \frac{7 2 ( \frac{3}{2} ) ^{\frac{5}{2}} - 45 3}{35 2}

+1

十進法表記

0.50925 \dots

解答ステップ

\int_{\frac{5 π}{6}}^{π} \frac{cos ^{4} ( x )}{1 - sin ( x )} d x

u置換積分法を適用する

= \int_{\frac{1}{2}}^{0} (u - 1) (u + 1)^{\frac{3}{2}} d u

\int_{a}^{b} f (x) d x = - \int_{b}^{a} f (x) d x, a < b

= - \int_{0}^{\frac{1}{2}} (u - 1) (u + 1)^{\frac{3}{2}} d u

拡張

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

= - \int_{0}^{\frac{1}{2}} u (u + 1)^{\frac{3}{2}} - (u + 1)^{\frac{3}{2}} 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

= - (\int_{0}^{\frac{1}{2}} u (u + 1)^{\frac{3}{2}} d u - \int_{0}^{\frac{1}{2}} (u + 1)^{\frac{3}{2}} d u)

\int_{0}^{\frac{1}{2}} u (u + 1)^{\frac{3}{2}} d u = \frac{14 2 ( \frac{3}{2} ) ^{\frac{5}{2}} - 27 3}{70 2} + \frac{4}{35}

\int_{0}^{\frac{1}{2}} (u + 1)^{\frac{3}{2}} d u = \frac{9 3}{10 2} - \frac{2}{5}

= - \frac{14 2 ( \frac{3}{2} ) ^{\frac{5}{2}} - 27 3}{70 2} + \frac{4}{35} - (\frac{9 3}{10 2} - \frac{2}{5})

簡素化

- \frac{14 2 ( \frac{3}{2} ) ^{\frac{5}{2}} - 27 3}{70 2} + \frac{4}{35} - (\frac{9 3}{10 2} - \frac{2}{5}) : - \frac{18}{35} - \frac{7 2 ( \frac{3}{2} ) ^{\frac{5}{2}} - 45 3}{35 2}

= - \frac{18}{35} - \frac{7 2 ( \frac{3}{2} ) ^{\frac{5}{2}} - 45 3}{35 2}

グラフ

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