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人気のある微分積分 >

integral of ((1-sin(x)))/(1+sin(x))

解

\int \frac{( 1 - sin ( x ) )}{1 + sin ( x )} d x

解

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

解答ステップ

\int \frac{1 - sin ( x )}{1 + sin ( x )} d x

u置換積分法を適用する

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

定数を除く:

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

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

以下の部分分数を得る:

\frac{1 + u ^{2} - 2 u}{( 1 + u ^{2} + 2 u ) ( 1 + u ^{2} )} : \frac{2}{( u + 1 ) ^{2}} - \frac{1}{u ^{2} + 1}

= 2 \cdot \int \frac{2}{( u + 1 ) ^{2}} - \frac{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

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

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

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

= 2 (- \frac{2}{u + 1} - arctan (u))

代用を戻す

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

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

定数を解答に追加する

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

グラフ

人気の例

integral of (x-8)

\int (x - 8) d x

integral of ((3x^2)/(1+x^2)+tan(2x))

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

integral of ((cos(x))^2)

\int ((cos (x))^{2}) d x

integral of 1/(10x^2+100x+250)

\int \frac{1}{10 x ^{2} + 100 x + 250} d x

integral of xe^{4x^2+3}

\int x e^{4 x^{2} + 3} d x