integral of sec^3(x)tan^2(x)

解

\int sec^{3} (x) tan^{2} (x) d x

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

解答ステップ

\int sec^{3} (x) tan^{2} (x) d x

三角関数の公式を使用して書き換える

= \int sec^{3} (x) (- 1 + sec^{2} (x)) d x

拡張

sec^{3} (x) (- 1 + sec^{2} (x)) : - sec^{3} (x) + sec^{5} (x)

= \int - sec^{3} (x) + sec^{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 sec^{3} (x) d x + \int sec^{5} (x) d x

\int sec^{3} (x) d x = \frac{1}{2} sec (x) tan (x) + \frac{1}{2} ln ∣ tan (x) + sec (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) ∣)

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

簡素化

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

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

定数を解答に追加する

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