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人気のある微分積分 >

integral of 1/(1-tan^2(x))

解

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

解

\frac{1}{2} arctan (tan (x)) + \frac{1}{4} ln ∣ tan (x) + 1 ∣ - \frac{1}{4} ln ∣ tan (x) - 1 ∣ + C

解答ステップ

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

因数

1 - tan^{2} (x) : - (- 1 + tan^{2} (x))

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

定数を除く:

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

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

u置換積分法を適用する

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

以下の部分分数を得る:

\frac{1}{( 1 + u ^{2} ) ( - 1 + u ^{2} )} : - \frac{1}{2 ( u ^{2} + 1 )} - \frac{1}{4 ( u + 1 )} + \frac{1}{4 ( u - 1 )}

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

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

\int \frac{1}{4 ( u + 1 )} d u = \frac{1}{4} ln ∣ u + 1 ∣

\int \frac{1}{4 ( u - 1 )} d u = \frac{1}{4} ln ∣ u - 1 ∣

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

代用を戻す

u = tan (x)

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

簡素化

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

定数を解答に追加する

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

グラフ

人気の例

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