integral of (sin(x)+1)/(cos(x))

解

\int \frac{sin ( x ) + 1}{cos ( x )} d x

- ln ∣ cos (x) ∣ + ln ∣ tan (x) + sec (x) ∣ + C

解答ステップ

\int \frac{sin ( x ) + 1}{cos ( x )} d x

拡張

\frac{sin ( x ) + 1}{cos ( x )} : \frac{sin ( x )}{cos ( x )} + \frac{1}{cos ( x )}

= \int \frac{sin ( x )}{cos ( x )} + \frac{1}{cos ( 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 \frac{sin ( x )}{cos ( x )} d x + \int \frac{1}{cos ( x )} d x

\int \frac{sin ( x )}{cos ( x )} d x = - ln ∣ cos (x) ∣

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

= - ln ∣ cos (x) ∣ + ln ∣ tan (x) + sec (x) ∣

定数を解答に追加する

= - ln ∣ cos (x) ∣ + ln ∣ tan (x) + sec (x) ∣ + C