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人気のある微分積分 >

integral of 1/(sin^5(x)*cos(x))

解

\int \frac{1}{sin ^{5} ( x ) \cdot cos ( x )} d x

解

- \frac{1}{4} cot^{4} (x) - cot^{2} (x) - ln ∣ cot (x) ∣ + C

解答ステップ

\int \frac{1}{sin ^{5} ( x ) cos ( x )} d x

三角関数による置換を適用する

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

定数を除く:

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

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

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

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

(u^{2} + 1) u^{2} + 1 u^{2} + 1 = (u^{2} + 1)^{2}

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

拡張

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

= - \int u^{3} + 2 u + \frac{1}{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

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

\int u^{3} d u = \frac{u ^{4}}{4}

\int 2 u d u = u^{2}

\int \frac{1}{u} d u = ln ∣ u ∣

= - (\frac{u ^{4}}{4} + u^{2} + ln ∣ u ∣)

代用を戻す

u = cot (x)

= - (\frac{cot ^{4} ( x )}{4} + cot^{2} (x) + ln ∣ cot (x) ∣)

簡素化

- (\frac{cot ^{4} ( x )}{4} + cot^{2} (x) + ln ∣ cot (x) ∣) : - \frac{1}{4} cot^{4} (x) - cot^{2} (x) - ln ∣ cot (x) ∣

= - \frac{1}{4} cot^{4} (x) - cot^{2} (x) - ln ∣ cot (x) ∣

定数を解答に追加する

= - \frac{1}{4} cot^{4} (x) - cot^{2} (x) - ln ∣ cot (x) ∣ + C

グラフ

人気の例

integral of (x^2)/(1+y^2)

\int \frac{x ^{2}}{1 + y ^{2}} d y

integral of (4x^3)/((2-x^4)^6)

\int \frac{4 x ^{3}}{( 2 - x ^{4} ) ^{6}} d x

integral of 1/10 e^{-1/10 x}

\int \frac{1}{10} e^{- \frac{1}{10} x} d x

integral of cos(ex)

\int cos (e x) d x

integral of (x^3-6x+5)/(x^2-x-6)

\int \frac{x ^{3} - 6 x + 5}{x ^{2} - x - 6} d x