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人気のある微分積分 >

integral of 1/(sin(x)-2csc(x))

解

\int \frac{1}{sin ( x ) - 2 csc ( x )} d x

解

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

解答ステップ

\int \frac{1}{sin ( x ) - 2 csc ( x )} d x

u置換積分法を適用する

= \int \frac{2 u}{- u ^{4} - 1} 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{u}{- u ^{4} - 1} d u

以下の部分分数を得る:

\frac{u}{- u ^{4} - 1} : \frac{1}{2 2 ( u ^{2} + 2 u + 1 )} - \frac{1}{2 2 ( u ^{2} - 2 u + 1 )}

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

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

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

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

代用を戻す

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

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

簡素化

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

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

定数を解答に追加する

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

グラフ

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