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

解答

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

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

求解步骤

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

使用三角恒等式改写

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

乘开

(- 1 + sec^{2} (x)) sec (x) : - sec (x) + sec^{3} (x)

= \int - sec (x) + sec^{3} (x) d x

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 sec (x) d x + \int sec^{3} (x) d x

\int sec (x) d x = ln ∣ tan (x) + sec (x) ∣

\int sec^{3} (x) d x = \frac{1}{2} sec (x) tan (x) + \frac{1}{2} ln ∣ tan (x) + sec (x) ∣

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

化简

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

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

解答补常数

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

\frac{d}{d x} (ln (ln (12 x)))

\int \frac{1}{2 sin ^{2} ( x ) + 3 cos ^{2} ( x )} d x

y^{^{''}} + 81 y = cos (2 t)

d er i v a t i v e y = \frac{2}{x ^{3}}

\frac{\partial}{\partial z} (x sin (y - z))