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人気のある微分積分 >

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

解

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

解

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

解答ステップ

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

u置換積分法を適用する

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

以下の部分分数を得る:

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

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

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

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

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

代用を戻す

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

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

簡素化

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

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

定数を解答に追加する

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

グラフ

人気の例

integral of sin(2x)e^{-2x}

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

integral of (sin(npix))

\int (sin (nπ x)) d x

integral of (sin(3t))/(cos(3t-1))

\int \frac{sin ( 3 t )}{cos ( 3 t - 1 )} d t

integral of 2x^2-2x+2

\int 2 x^{2} - 2 x + 2 d x

integral of (sqrt(2x^2-16))/(3x)

\int \frac{2 x ^{2} - 16}{3 x} d x