integral of (sec(t)-1)^2

解

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

tan (t) - 2 ln ∣ tan (t) + sec (t) ∣ + t + C

解答ステップ

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

三角関数の公式を使用して書き換える

= \int \frac{( 1 - cos ( t ) ) ^{2}}{cos ^{2} ( t )} d t

拡張

\frac{( 1 - cos ( t ) ) ^{2}}{cos ^{2} ( t )} : \frac{1}{cos ^{2} ( t )} - \frac{2}{cos ( t )} + 1

= \int \frac{1}{cos ^{2} ( t )} - \frac{2}{cos ( t )} + 1 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 \frac{1}{cos ^{2} ( t )} d t - \int \frac{2}{cos ( t )} d t + \int 1 d t

\int \frac{1}{cos ^{2} ( t )} d t = tan (t)

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

\int 1 d t = t

= tan (t) - 2 ln ∣ tan (t) + sec (t) ∣ + t

定数を解答に追加する

= tan (t) - 2 ln ∣ tan (t) + sec (t) ∣ + t + C