integral of 1/(sec(x)+1)

解

\int \frac{1}{sec ( x ) + 1} d x

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

解答ステップ

\int \frac{1}{sec ( x ) + 1} d x

u置換積分法を適用する

= \int \frac{1 - u ^{2}}{1 + u ^{2}} d u

\frac{1 - u ^{2}}{1 + u ^{2}} = \frac{2}{1 + u ^{2}} - 1

= \int \frac{2}{1 + u ^{2}} - 1 d u

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{2}{1 + u ^{2}} d u - \int 1 d u

\int \frac{2}{1 + u ^{2}} d u = 2 arctan (u)

\int 1 d u = u

= 2 arctan (u) - u

代用を戻す

u = tan (\frac{x}{2})

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

定数を解答に追加する

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