integral of 1/(cos^2(x)+5cos(x)+6)

解

\int \frac{1}{cos ^{2} ( x ) + 5 cos ( x ) + 6} d x

\frac{2}{3} arctan (\frac{1}{3} tan (\frac{x}{2})) - \frac{1}{2} arctan (\frac{tan ( \frac{x}{2} )}{2}) + C

解答ステップ

\int \frac{1}{cos ^{2} ( x ) + 5 cos ( x ) + 6} d x

以下の部分分数を得る:

\frac{1}{cos ^{2} ( x ) + 5 cos ( x ) + 6} : \frac{1}{cos ( x ) + 2} - \frac{1}{cos ( x ) + 3}

= \int \frac{1}{cos ( x ) + 2} - \frac{1}{cos ( x ) + 3} 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{1}{cos ( x ) + 2} d x - \int \frac{1}{cos ( x ) + 3} d x

\int \frac{1}{cos ( x ) + 2} d x = \frac{2}{3} arctan (\frac{1}{3} tan (\frac{x}{2}))

\int \frac{1}{cos ( x ) + 3} d x = \frac{1}{2} arctan (\frac{tan ( \frac{x}{2} )}{2})

= \frac{2}{3} arctan (\frac{1}{3} tan (\frac{x}{2})) - \frac{1}{2} arctan (\frac{tan ( \frac{x}{2} )}{2})

定数を解答に追加する

= \frac{2}{3} arctan (\frac{1}{3} tan (\frac{x}{2})) - \frac{1}{2} arctan (\frac{tan ( \frac{x}{2} )}{2}) + C