integral of (1+cos(x))/((1+sin(x))^2)

解

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

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

解答ステップ

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

拡張

\frac{1 + cos ( x )}{( 1 + sin ( x ) ) ^{2}} : \frac{1}{( 1 + sin ( x ) ) ^{2}} + \frac{cos ( x )}{( 1 + sin ( x ) ) ^{2}}

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

\int \frac{1}{( 1 + sin ( x ) ) ^{2}} d x = 2 (- \frac{1}{tan ( \frac{x}{2} ) + 1} + \frac{1}{( tan ( \frac{x}{2} ) + 1 ) ^{2}} - \frac{2}{3 ( tan ( \frac{x}{2} ) + 1 ) ^{3}})

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

= 2 (- \frac{1}{tan ( \frac{x}{2} ) + 1} + \frac{1}{( tan ( \frac{x}{2} ) + 1 ) ^{2}} - \frac{2}{3 ( tan ( \frac{x}{2} ) + 1 ) ^{3}}) - \frac{1}{1 + sin ( x )}

定数を解答に追加する

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