integral of (tan^3(x)+tan^5(x))

解

\int (tan^{3} (x) + tan^{5} (x)) d x

\frac{1}{2} sec^{2} (x) - sec^{2} (x) + \frac{1}{4} sec^{4} (x) + C

解答ステップ

\int tan^{3} (x) + tan^{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^{3} (x) d x + \int tan^{5} (x) d x

\int tan^{3} (x) d x = - ln ∣ sec (x) ∣ + \frac{sec ^{2} ( x )}{2}

\int tan^{5} (x) d x = ln ∣ sec (x) ∣ - sec^{2} (x) + \frac{sec ^{4} ( x )}{4}

= - ln ∣ sec (x) ∣ + \frac{sec ^{2} ( x )}{2} + ln ∣ sec (x) ∣ - sec^{2} (x) + \frac{sec ^{4} ( x )}{4}

簡素化

- ln ∣ sec (x) ∣ + \frac{sec ^{2} ( x )}{2} + ln ∣ sec (x) ∣ - sec^{2} (x) + \frac{sec ^{4} ( x )}{4} : \frac{1}{2} sec^{2} (x) - sec^{2} (x) + \frac{1}{4} sec^{4} (x)

= \frac{1}{2} sec^{2} (x) - sec^{2} (x) + \frac{1}{4} sec^{4} (x)

定数を解答に追加する

= \frac{1}{2} sec^{2} (x) - sec^{2} (x) + \frac{1}{4} sec^{4} (x) + C