integral of sec^9(x)

解

\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

解答ステップ

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

積分換算を適用する

= \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) ∣)))

定数を解答に追加する

= \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