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人気のある微分積分 >

integral of cot^4(1/2 x)

解

\int cot^{4} (\frac{1}{2} x) d x

解

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

解答ステップ

\int cot^{4} (\frac{1}{2} x) d x

三角関数の公式を使用して書き換える

= \int \frac{1}{tan ^{4} ( \frac{x}{2} )} d x

基本的な三角関数の公式を使用する:

\frac{1}{tan ( x )} = cot (x)

= \int cot^{4} (\frac{x}{2}) d x

u置換積分法を適用する

= \int \frac{2}{u ^{4} ( 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 ^{4} ( 1 + u ^{2} )} d u

以下の部分分数を得る:

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

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

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

\int \frac{1}{u ^{4}} d u = - \frac{1}{3 u ^{3}}

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

= 2 (- (- \frac{1}{u}) - \frac{1}{3 u ^{3}} + arctan (u))

代用を戻す

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

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

簡素化

2 (- (- \frac{1}{tan ( \frac{x}{2} )}) - \frac{1}{3 tan ^{3} ( \frac{x}{2} )} + arctan (tan (\frac{x}{2}))) : 2 (cot (\frac{x}{2}) - \frac{1}{3} cot^{3} (\frac{x}{2}) + arctan (tan (\frac{x}{2})))

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

定数を解答に追加する

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

グラフ

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