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

解

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

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

解答ステップ

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

三角関数の公式を使用して書き換える

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

拡張

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

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

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} (t) d t + \int sec^{5} (t) d t

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

\int sec^{5} (t) d t = \frac{sec ^{4} ( t ) sin ( t )}{4} + \frac{3}{4} (\frac{1}{2} sec (t) tan (t) + \frac{1}{2} ln ∣ tan (t) + sec (t) ∣)

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

簡素化

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

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

定数を解答に追加する

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