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人気のある微分積分 >

integral of (cot(5x)+csc(5x))^2

解

\int (cot (5 x) + csc (5 x))^{2} d x

解

\frac{2}{5} (- cot (\frac{5 x}{2}) - arctan (tan (\frac{5 x}{2}))) + C

解答ステップ

\int (cot (5 x) + csc (5 x))^{2} d x

u置換積分法を適用する

= \int (cot (u) + csc (u))^{2} \frac{1}{5} d u

定数を除く:

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

= \frac{1}{5} \cdot \int (cot (u) + csc (u))^{2} d u

u置換積分法を適用する

= \frac{1}{5} \cdot \int \frac{2}{v ^{2} ( 1 + v ^{2} )} d v

定数を除く:

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

= \frac{1}{5} \cdot 2 \cdot \int \frac{1}{v ^{2} ( 1 + v ^{2} )} d v

以下の部分分数を得る:

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

= \frac{1}{5} \cdot 2 \cdot \int \frac{1}{v ^{2}} - \frac{1}{v ^{2} + 1} d v

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

= \frac{1}{5} \cdot 2 (\int \frac{1}{v ^{2}} d v - \int \frac{1}{v ^{2} + 1} d v)

\int \frac{1}{v ^{2}} d v = - \frac{1}{v}

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

= \frac{1}{5} \cdot 2 (- \frac{1}{v} - arctan (v))

代用を戻す

= \frac{1}{5} \cdot 2 (- \frac{1}{tan ( \frac{5 x}{2} )} - arctan (tan (\frac{5 x}{2})))

簡素化

\frac{1}{5} \cdot 2 (- \frac{1}{tan ( \frac{5 x}{2} )} - arctan (tan (\frac{5 x}{2}))) : \frac{2}{5} (- cot (\frac{5 x}{2}) - arctan (tan (\frac{5 x}{2})))

= \frac{2}{5} (- cot (\frac{5 x}{2}) - arctan (tan (\frac{5 x}{2})))

定数を解答に追加する

= \frac{2}{5} (- cot (\frac{5 x}{2}) - arctan (tan (\frac{5 x}{2}))) + C

グラフ

人気の例

integral of (x^4-xarctan(x))/((1+x^2)^2)

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