integral of 19tan^2(x)sec^3(x)

解答

\int 19 tan^{2} (x) sec^{3} (x) d x

19 (- \frac{1}{4} (\frac{1}{2} sec (x) tan (x) + \frac{1}{2} ln ∣ tan (x) + sec (x) ∣) + \frac{1}{4} sec^{4} (x) sin (x)) + C

求解步骤

\int 19 tan^{2} (x) sec^{3} (x) d x

提出常数:

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

= 19 \cdot \int tan^{2} (x) sec^{3} (x) d x

使用三角恒等式改写

= 19 \cdot \int (- 1 + sec^{2} (x)) sec^{3} (x) d x

乘开

(- 1 + sec^{2} (x)) sec^{3} (x) : - sec^{3} (x) + sec^{5} (x)

= 19 \cdot \int - sec^{3} (x) + sec^{5} (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

= 19 (- \int sec^{3} (x) d x + \int sec^{5} (x) d x)

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

\int sec^{5} (x) d x = \frac{sec ^{4} ( x ) sin ( x )}{4} + \frac{3}{4} (\frac{1}{2} sec (x) tan (x) + \frac{1}{2} ln ∣ tan (x) + sec (x) ∣)

= 19 (- (\frac{1}{2} sec (x) tan (x) + \frac{1}{2} ln ∣ tan (x) + sec (x) ∣) + \frac{sec ^{4} ( x ) sin ( x )}{4} + \frac{3}{4} (\frac{1}{2} sec (x) tan (x) + \frac{1}{2} ln ∣ tan (x) + sec (x) ∣))

化简

19 (- (\frac{1}{2} sec (x) tan (x) + \frac{1}{2} ln ∣ tan (x) + sec (x) ∣) + \frac{sec ^{4} ( x ) sin ( x )}{4} + \frac{3}{4} (\frac{1}{2} sec (x) tan (x) + \frac{1}{2} ln ∣ tan (x) + sec (x) ∣)) : 19 (- \frac{1}{4} (\frac{1}{2} sec (x) tan (x) + \frac{1}{2} ln ∣ tan (x) + sec (x) ∣) + \frac{1}{4} sec^{4} (x) sin (x))

= 19 (- \frac{1}{4} (\frac{1}{2} sec (x) tan (x) + \frac{1}{2} ln ∣ tan (x) + sec (x) ∣) + \frac{1}{4} sec^{4} (x) sin (x))

解答补常数

= 19 (- \frac{1}{4} (\frac{1}{2} sec (x) tan (x) + \frac{1}{2} ln ∣ tan (x) + sec (x) ∣) + \frac{1}{4} sec^{4} (x) sin (x)) + C