integral of tan^2(x)sec^5(x)

解

\int tan^{2} (x) sec^{5} (x) d x

- \frac{1}{16} sec (x) tan (x) - \frac{1}{16} ln ∣ tan (x) + sec (x) ∣ + \frac{1}{6} sec^{6} (x) sin (x) + \frac{5}{24} sec^{4} (x) sin (x) - \frac{1}{4} sec^{4} (x) sin (x) + C

解答ステップ

\int tan^{2} (x) sec^{5} (x) d x

三角関数の公式を使用して書き換える

= \int (- 1 + sec^{2} (x)) sec^{5} (x) d x

拡張

(- 1 + sec^{2} (x)) sec^{5} (x) : - sec^{5} (x) + sec^{7} (x)

= \int - sec^{5} (x) + sec^{7} (x) d x

総和規則を適用する:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= - \int sec^{5} (x) d x + \int sec^{7} (x) d 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) ∣)

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

簡素化

- (\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) ∣)) + \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{1}{16} sec (x) tan (x) - \frac{1}{16} ln ∣ tan (x) + sec (x) ∣ + \frac{1}{6} sec^{6} (x) sin (x) + \frac{5}{24} sec^{4} (x) sin (x) - \frac{1}{4} sec^{4} (x) sin (x)

= - \frac{1}{16} sec (x) tan (x) - \frac{1}{16} ln ∣ tan (x) + sec (x) ∣ + \frac{1}{6} sec^{6} (x) sin (x) + \frac{5}{24} sec^{4} (x) sin (x) - \frac{1}{4} sec^{4} (x) sin (x)

定数を解答に追加する

= - \frac{1}{16} sec (x) tan (x) - \frac{1}{16} ln ∣ tan (x) + sec (x) ∣ + \frac{1}{6} sec^{6} (x) sin (x) + \frac{5}{24} sec^{4} (x) sin (x) - \frac{1}{4} sec^{4} (x) sin (x) + C