integral of (cos(3x)+3)/(sin(3x))

解

\int \frac{cos ( 3 x ) + 3}{sin ( 3 x )} d x

\frac{1}{3} ln ∣ sin (3 x) ∣ + ln tan (\frac{3 x}{2}) + C

解答ステップ

\int \frac{cos ( 3 x ) + 3}{sin ( 3 x )} d x

拡張

\frac{cos ( 3 x ) + 3}{sin ( 3 x )} : \frac{cos ( 3 x )}{sin ( 3 x )} + \frac{3}{sin ( 3 x )}

= \int \frac{cos ( 3 x )}{sin ( 3 x )} + \frac{3}{sin ( 3 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{cos ( 3 x )}{sin ( 3 x )} d x + \int \frac{3}{sin ( 3 x )} d x

\int \frac{cos ( 3 x )}{sin ( 3 x )} d x = \frac{1}{3} ln ∣ sin (3 x) ∣

\int \frac{3}{sin ( 3 x )} d x = ln tan (\frac{3 x}{2})

= \frac{1}{3} ln ∣ sin (3 x) ∣ + ln tan (\frac{3 x}{2})

定数を解答に追加する

= \frac{1}{3} ln ∣ sin (3 x) ∣ + ln tan (\frac{3 x}{2}) + C