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人気のある微分積分 >

integral of (cos(x))/(sin^2(x)-2)

解

\int \frac{cos ( x )}{sin ^{2} ( x ) - 2} d x

解

- \frac{1}{2 2} (ln \frac{sin ( x )}{2} + 1 - ln \frac{sin ( x )}{2} - 1) + C

解答ステップ

\int \frac{cos ( x )}{sin ^{2} ( x ) - 2} d x

u置換積分法を適用する

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

置換積分法を適用する

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

定数を除く:

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

= \frac{1}{2} \cdot \int \frac{1}{v ^{2} - 1} d v

因数

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

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

定数を除く:

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

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

共通積分を使用する:

\int \frac{1}{- v ^{2} + 1} d v = \frac{ln ∣ v + 1 ∣}{2} - \frac{ln ∣ v - 1 ∣}{2}

= \frac{1}{2} (- (\frac{ln ∣ v + 1 ∣}{2} - \frac{ln ∣ v - 1 ∣}{2}))

代用を戻す

= \frac{1}{2} - \frac{ln \frac{s i n ( x )}{2} + 1}{2} - \frac{ln \frac{s i n ( x )}{2} - 1}{2}

簡素化

\frac{1}{2} - \frac{ln \frac{s i n ( x )}{2} + 1}{2} - \frac{ln \frac{s i n ( x )}{2} - 1}{2} : - \frac{1}{2 2} (ln \frac{sin ( x )}{2} + 1 - ln \frac{sin ( x )}{2} - 1)

= - \frac{1}{2 2} (ln \frac{sin ( x )}{2} + 1 - ln \frac{sin ( x )}{2} - 1)

定数を解答に追加する

= - \frac{1}{2 2} (ln \frac{sin ( x )}{2} + 1 - ln \frac{sin ( x )}{2} - 1) + C

グラフ

人気の例

sum from n=1 to infinity of 1/((n^2+1))

n = 1 \sum \infty \frac{1}{( n ^{2} + 1 )}

limit as x approaches 0+of x^{-(17)/(ln(x))}

x \to 0 + lim (x^{- \frac{17}{l n ( x )}})

(dy)/(dx)=3x(e^{2y})

\frac{d y}{d x} = 3 x (e^{2 y})

derivative of (3x+3/(4e^x))

\frac{d}{d x} (\frac{3 x + 3}{4 e ^{x}})

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

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