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

解答

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

- \frac{1}{2} ln ∣ cos (2 x) ∣ + C

求解步骤

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

提出常数:

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

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

使用三角恒等式改写

= 2 \cdot \int \frac{sin ( x ) cos ( x )}{cos ^{2} ( x ) - sin ^{2} ( x )} d x

使用分布积分法

= 2 (x cos (x) sin (x) sec (2 x) - \int x sec^{2} (2 x) d x)

\int x sec^{2} (2 x) d x = \frac{1}{2} x tan (2 x) + \frac{1}{4} ln ∣ cos (2 x) ∣

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

化简

2 (x cos (x) sin (x) sec (2 x) - (\frac{1}{2} x tan (2 x) + \frac{1}{4} ln ∣ cos (2 x) ∣)) : - \frac{1}{2} ln ∣ cos (2 x) ∣

= - \frac{1}{2} ln ∣ cos (2 x) ∣

解答补常数

= - \frac{1}{2} ln ∣ cos (2 x) ∣ + C

\frac{d}{d x} (tan (2 x + 1))

f (x) = \frac{2 x}{x + 1}

\frac{d}{d x} (\frac{x + 2}{x ^{2} - 4})

\frac{d}{d x} (\frac{6 x ^{2} + 2 x + 8}{x})

t an g e n t y = 3 x^{3} - x^{2} + 8, (2, 28)