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

解

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

- x + \frac{2}{5} tan (5 x) - \frac{2}{5} sec (5 x) + C

解答ステップ

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

拡張

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

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

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

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

\int sec^{2} (5 x) d x = \frac{1}{5} tan (5 x)

= - x + \frac{1}{5} tan (5 x) - \frac{2}{5} sec (5 x) + \frac{1}{5} tan (5 x)

類似した元を足す:

\frac{1}{5} tan (5 x) + \frac{1}{5} tan (5 x) = \frac{2}{5} tan (5 x)

= - x + \frac{2}{5} tan (5 x) - \frac{2}{5} sec (5 x)

定数を解答に追加する

= - x + \frac{2}{5} tan (5 x) - \frac{2}{5} sec (5 x) + C