integral of sec^9(x)

解答

\int sec^{9} (x) d x

\frac{sec ^{8} ( x ) sin ( x )}{8} + \frac{7}{8} (\frac{1}{6} sec^{6} (x) sin (x) + \frac{5}{48} (2 sec^{4} (x) sin (x) + 3 (sec (x) tan (x) + ln ∣ tan (x) + sec (x) ∣))) + C

求解步骤

\int sec^{9} (x) d x

使用降次积分法

= \frac{sec ^{8} ( x ) sin ( x )}{8} + \frac{7}{8} \cdot \int sec^{7} (x) d x

\int sec^{7} (x) d x = \frac{1}{6} sec^{6} (x) sin (x) + \frac{5}{48} (2 sec^{4} (x) sin (x) + 3 (sec (x) tan (x) + ln ∣ tan (x) + sec (x) ∣))

= \frac{sec ^{8} ( x ) sin ( x )}{8} + \frac{7}{8} (\frac{1}{6} sec^{6} (x) sin (x) + \frac{5}{48} (2 sec^{4} (x) sin (x) + 3 (sec (x) tan (x) + ln ∣ tan (x) + sec (x) ∣)))

解答补常数

= \frac{sec ^{8} ( x ) sin ( x )}{8} + \frac{7}{8} (\frac{1}{6} sec^{6} (x) sin (x) + \frac{5}{48} (2 sec^{4} (x) sin (x) + 3 (sec (x) tan (x) + ln ∣ tan (x) + sec (x) ∣))) + C

\int_{0}^{1} ln (t) d t

\frac{\partial}{\partial x} (2 x^{\frac{1}{2}} + y)

d er i v a t i v e f (x) = 9 x + 7

\int \frac{3}{6 + 3 x} d x

\int \frac{2}{nπ} sin (\frac{nπ x}{2}) d x