integral of sec(t)(9sec(t)+7tan(t))

解

\int sec (t) (9 sec (t) + 7 tan (t)) d t

9 tan (t) + 7 sec (t) + C

解答ステップ

\int sec (t) (9 sec (t) + 7 tan (t)) d t

三角関数の公式を使用して書き換える

= \int \frac{9 + 7 sin ( t )}{cos ^{2} ( t )} d t

拡張

\frac{9 + 7 sin ( t )}{cos ^{2} ( t )} : \frac{9}{cos ^{2} ( t )} + \frac{7 sin ( t )}{cos ^{2} ( t )}

= \int \frac{9}{cos ^{2} ( t )} + \frac{7 sin ( t )}{cos ^{2} ( 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 \frac{9}{cos ^{2} ( t )} d t + \int \frac{7 sin ( t )}{cos ^{2} ( t )} d t

\int \frac{9}{cos ^{2} ( t )} d t = 9 tan (t)

\int \frac{7 sin ( t )}{cos ^{2} ( t )} d t = 7 sec (t)

= 9 tan (t) + 7 sec (t)

定数を解答に追加する

= 9 tan (t) + 7 sec (t) + C