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人気のある微分積分 >

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

解

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

解

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

解答ステップ

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

u置換積分法を適用する

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

以下の部分分数を得る:

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

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

\int \frac{- u - 1}{2 ( u ^{2} + 2 u - 1 )} d u = - \frac{1}{4} ln u^{2} + 2 u - 1

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

= 2 (- \frac{1}{4} ln u^{2} + 2 u - 1 + \frac{1}{2} (\frac{1}{2} ln u^{2} + 1 + arctan (u)))

代用を戻す

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

= 2 (- \frac{1}{4} ln tan^{2} (\frac{x}{2}) + 2 tan (\frac{x}{2}) - 1 + \frac{1}{2} (\frac{1}{2} ln tan^{2} (\frac{x}{2}) + 1 + arctan (tan (\frac{x}{2}))))

簡素化

2 (- \frac{1}{4} ln tan^{2} (\frac{x}{2}) + 2 tan (\frac{x}{2}) - 1 + \frac{1}{2} (\frac{1}{2} ln tan^{2} (\frac{x}{2}) + 1 + arctan (tan (\frac{x}{2})))) : - \frac{1}{2} ln tan^{2} (\frac{x}{2}) + 2 tan (\frac{x}{2}) - 1 + \frac{1}{2} ln (sec^{2} (\frac{x}{2})) + arctan (tan (\frac{x}{2}))

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

定数を解答に追加する

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

グラフ

人気の例

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