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受欢迎的微积分 >

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

解答

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

解答

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

求解步骤

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

因式分解

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

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

提出常数:

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

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

使用换元积分法

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

将

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

用部份分式展开:

- \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

使用积分加法定则:

\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 ∣)

u = tan (x)

代回

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

化简

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

解答补常数

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

作图

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