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人気のある微分積分 >

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

解

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

解

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

解答ステップ

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

因数

cos (x) - sin (x) - 1 : - (- cos (x) + sin (x) + 1)

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

定数を除く:

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

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

u置換積分法を適用する

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

以下の部分分数を得る:

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

= - \int \frac{1}{u} - \frac{1}{u + 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

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

\int \frac{1}{u} d u = ln ∣ u ∣

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

= - (ln ∣ u ∣ - ln ∣ u + 1 ∣)

代用を戻す

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

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

簡素化

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

定数を解答に追加する

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

グラフ

人気の例

integral of 1-(2x)/pi

\int 1 - \frac{2 x}{π} d x

integral of (cos(θ))/(9sin^2(θ))

\int \frac{cos ( θ )}{9 sin ^{2} ( θ )} d θ

integral of (tan^3(x))(sec^2(x))

\int (tan^{3} (x)) (sec^{2} (x)) d x

integral of (sqrt(x)+1/(\sqrt[3]{x)})^3

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

integral of-3sin(2x+1)

\int - 3 sin (2 x + 1) d x