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人気のある微分積分 >

integral of (tan(x))/(1-cos(x))

解

\int \frac{tan ( x )}{1 - cos ( x )} d x

解

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

解答ステップ

\int \frac{tan ( x )}{1 - cos ( x )} d x

u置換積分法を適用する

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

定数を除く:

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

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

u置換積分法を適用する

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

定数を除く:

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

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

以下の部分分数を得る:

\frac{1}{v ( - v + 1 )} : \frac{1}{v} - \frac{1}{v - 1}

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

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

\int \frac{1}{v} d v = ln ∣ v ∣

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

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

代用を戻す

= 2 \cdot \frac{1}{2} (ln tan^{2} (\frac{x}{2}) - ln tan^{2} (\frac{x}{2}) - 1)

簡素化

2 \cdot \frac{1}{2} (ln tan^{2} (\frac{x}{2}) - ln tan^{2} (\frac{x}{2}) - 1) : ln (tan^{2} (\frac{x}{2})) - ln tan^{2} (\frac{x}{2}) - 1

= ln (tan^{2} (\frac{x}{2})) - ln tan^{2} (\frac{x}{2}) - 1

定数を解答に追加する

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

グラフ

人気の例

integral of (1/(3x^2))

\int (\frac{1}{3 x ^{2}}) d x

integral of x^6)

\int d (x^{6}) d x

integral of (2x^2+4)^8

\int (2 x^{2} + 4)^{8} d x

integral of 5x^4e^{5x^5}

\int 5 x^{4} e^{5 x^{5}} d x

integral of (x/(sqrt(x^2-4)))

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