integral of 5(cos(x)+sin(x))/(sin(2x))

解

\int 5 \frac{cos ( x ) + sin ( x )}{sin ( 2 x )} d x

5 (\frac{1}{2} ln tan (\frac{x}{2}) + \frac{1}{2} ln ∣ tan (x) + sec (x) ∣) + C

解答ステップ

\int 5 \cdot \frac{cos ( x ) + sin ( x )}{sin ( 2 x )} d x

定数を除く:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 5 \cdot \int \frac{cos ( x ) + sin ( x )}{sin ( 2 x )} d x

拡張

\frac{cos ( x ) + sin ( x )}{sin ( 2 x )} : \frac{cos ( x )}{sin ( 2 x )} + \frac{sin ( x )}{sin ( 2 x )}

= 5 \cdot \int \frac{cos ( x )}{sin ( 2 x )} + \frac{sin ( x )}{sin ( 2 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

= 5 (\int \frac{cos ( x )}{sin ( 2 x )} d x + \int \frac{sin ( x )}{sin ( 2 x )} d x)

\int \frac{cos ( x )}{sin ( 2 x )} d x = \frac{1}{2} ln tan (\frac{x}{2})

\int \frac{sin ( x )}{sin ( 2 x )} d x = \frac{1}{2} ln ∣ tan (x) + sec (x) ∣

= 5 (\frac{1}{2} ln tan (\frac{x}{2}) + \frac{1}{2} ln ∣ tan (x) + sec (x) ∣)

定数を解答に追加する

= 5 (\frac{1}{2} ln tan (\frac{x}{2}) + \frac{1}{2} ln ∣ tan (x) + sec (x) ∣) + C