integral of 5tan^2(x)+tan^4(x)

解

\int 5 tan^{2} (x) + tan^{4} (x) d x

- 4 x + \frac{1}{3} tan^{3} (x) + 4 tan (x) + C

解答ステップ

\int 5 tan^{2} (x) + tan^{4} (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 5 tan^{2} (x) d x + \int tan^{4} (x) d x

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

\int tan^{4} (x) d x = \frac{1}{3} tan^{3} (x) + x - tan (x)

= 5 (- x + tan (x)) + \frac{1}{3} tan^{3} (x) + x - tan (x)

簡素化

5 (- x + tan (x)) + \frac{1}{3} tan^{3} (x) + x - tan (x) : - 4 x + \frac{1}{3} tan^{3} (x) + 4 tan (x)

= - 4 x + \frac{1}{3} tan^{3} (x) + 4 tan (x)

定数を解答に追加する

= - 4 x + \frac{1}{3} tan^{3} (x) + 4 tan (x) + C