integral of (4-sec^2(3x))/(tan(3x))

解

\int \frac{4 - sec ^{2} ( 3 x )}{tan ( 3 x )} d x

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

解答ステップ

\int \frac{4 - sec ^{2} ( 3 x )}{tan ( 3 x )} d x

拡張

\frac{4 - sec ^{2} ( 3 x )}{tan ( 3 x )} : \frac{4}{tan ( 3 x )} - \frac{sec ^{2} ( 3 x )}{tan ( 3 x )}

= \int \frac{4}{tan ( 3 x )} - \frac{sec ^{2} ( 3 x )}{tan ( 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{4}{tan ( 3 x )} d x - \int \frac{sec ^{2} ( 3 x )}{tan ( 3 x )} d x

\int \frac{4}{tan ( 3 x )} d x = \frac{4}{3} ln ∣ sin (3 x) ∣

\int \frac{sec ^{2} ( 3 x )}{tan ( 3 x )} d x = \frac{1}{3} ln ∣ sin (3 x) ∣ - \frac{1}{3} ln ∣ cos (3 x) ∣

= \frac{4}{3} ln ∣ sin (3 x) ∣ - (\frac{1}{3} ln ∣ sin (3 x) ∣ - \frac{1}{3} ln ∣ cos (3 x) ∣)

簡素化

\frac{4}{3} ln ∣ sin (3 x) ∣ - (\frac{1}{3} ln ∣ sin (3 x) ∣ - \frac{1}{3} ln ∣ cos (3 x) ∣) : ln ∣ sin (3 x) ∣ + \frac{1}{3} ln ∣ cos (3 x) ∣

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

定数を解答に追加する

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