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人気のある微分積分 >

integral of sqrt((1+cos(x))/(1-cos(x)))

解

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

解

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

解答ステップ

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

u置換積分法を適用する

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

以下の部分分数を得る:

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

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

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

\int \frac{1}{u} d u = ln ∣ u ∣

\int \frac{u}{u ^{2} + 1} d u = \frac{1}{2} ln u^{2} + 1

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

代用を戻す

u = tan (\frac{x}{2})

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

簡素化

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

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

定数を解答に追加する

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

グラフ

人気の例

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\frac{\partial}{\partial x} ((3 x^{2} + y) e^{x y^{2}})