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人気のある微分積分 >

integral of tan^4(x)sec(x)

解

\int tan^{4} (x) sec (x) d x

解

32 (\frac{3}{256} ln tan (\frac{x}{2}) + 1 - \frac{3}{256 ( tan ( \frac{x}{2} ) + 1 )} + \frac{1}{256 ( tan ( \frac{x}{2} ) + 1 ) ^{2}} + \frac{1}{64 ( tan ( \frac{x}{2} ) + 1 ) ^{3}} - \frac{1}{128 ( tan ( \frac{x}{2} ) + 1 ) ^{4}} - \frac{3}{256} ln tan (\frac{x}{2}) - 1 - \frac{3}{256 ( tan ( \frac{x}{2} ) - 1 )} - \frac{1}{256 ( tan ( \frac{x}{2} ) - 1 ) ^{2}} + \frac{1}{64 ( tan ( \frac{x}{2} ) - 1 ) ^{3}} + \frac{1}{128 ( tan ( \frac{x}{2} ) - 1 ) ^{4}}) + C

解答ステップ

\int tan^{4} (x) sec (x) d x

三角関数の公式を使用して書き換える

= \int \frac{sin ^{4} ( x )}{cos ^{5} ( x )} d x

u置換積分法を適用する

= \int \frac{32 u ^{4}}{( 1 - u ^{2} ) ^{5}} d u

定数を除く:

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

= 32 \cdot \int \frac{u ^{4}}{( 1 - u ^{2} ) ^{5}} d u

以下の部分分数を得る:

\frac{u ^{4}}{( 1 - u ^{2} ) ^{5}} : \frac{3}{256 ( u + 1 )} + \frac{3}{256 ( u + 1 ) ^{2}} - \frac{1}{128 ( u + 1 ) ^{3}} - \frac{3}{64 ( u + 1 ) ^{4}} + \frac{1}{32 ( u + 1 ) ^{5}} - \frac{3}{256 ( u - 1 )} + \frac{3}{256 ( u - 1 ) ^{2}} + \frac{1}{128 ( u - 1 ) ^{3}} - \frac{3}{64 ( u - 1 ) ^{4}} - \frac{1}{32 ( u - 1 ) ^{5}}

= 32 \cdot \int \frac{3}{256 ( u + 1 )} + \frac{3}{256 ( u + 1 ) ^{2}} - \frac{1}{128 ( u + 1 ) ^{3}} - \frac{3}{64 ( u + 1 ) ^{4}} + \frac{1}{32 ( u + 1 ) ^{5}} - \frac{3}{256 ( u - 1 )} + \frac{3}{256 ( u - 1 ) ^{2}} + \frac{1}{128 ( u - 1 ) ^{3}} - \frac{3}{64 ( u - 1 ) ^{4}} - \frac{1}{32 ( u - 1 ) ^{5}} 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

= 32 (\int \frac{3}{256 ( u + 1 )} d u + \int \frac{3}{256 ( u + 1 ) ^{2}} d u - \int \frac{1}{128 ( u + 1 ) ^{3}} d u - \int \frac{3}{64 ( u + 1 ) ^{4}} d u + \int \frac{1}{32 ( u + 1 ) ^{5}} d u - \int \frac{3}{256 ( u - 1 )} d u + \int \frac{3}{256 ( u - 1 ) ^{2}} d u + \int \frac{1}{128 ( u - 1 ) ^{3}} d u - \int \frac{3}{64 ( u - 1 ) ^{4}} d u - \int \frac{1}{32 ( u - 1 ) ^{5}} d u)

\int \frac{3}{256 ( u + 1 )} d u = \frac{3}{256} ln ∣ u + 1 ∣

\int \frac{3}{256 ( u + 1 ) ^{2}} d u = - \frac{3}{256 ( u + 1 )}

\int \frac{1}{128 ( u + 1 ) ^{3}} d u = - \frac{1}{256 ( u + 1 ) ^{2}}

\int \frac{3}{64 ( u + 1 ) ^{4}} d u = - \frac{1}{64 ( u + 1 ) ^{3}}

\int \frac{1}{32 ( u + 1 ) ^{5}} d u = - \frac{1}{128 ( u + 1 ) ^{4}}

\int \frac{3}{256 ( u - 1 )} d u = \frac{3}{256} ln ∣ u - 1 ∣

\int \frac{3}{256 ( u - 1 ) ^{2}} d u = - \frac{3}{256 ( u - 1 )}

\int \frac{1}{128 ( u - 1 ) ^{3}} d u = - \frac{1}{256 ( u - 1 ) ^{2}}

\int \frac{3}{64 ( u - 1 ) ^{4}} d u = - \frac{1}{64 ( u - 1 ) ^{3}}

\int \frac{1}{32 ( u - 1 ) ^{5}} d u = - \frac{1}{128 ( u - 1 ) ^{4}}

= 32 (\frac{3}{256} ln ∣ u + 1 ∣ - \frac{3}{256 ( u + 1 )} - (- \frac{1}{256 ( u + 1 ) ^{2}}) - (- \frac{1}{64 ( u + 1 ) ^{3}}) - \frac{1}{128 ( u + 1 ) ^{4}} - \frac{3}{256} ln ∣ u - 1 ∣ - \frac{3}{256 ( u - 1 )} - \frac{1}{256 ( u - 1 ) ^{2}} - (- \frac{1}{64 ( u - 1 ) ^{3}}) - (- \frac{1}{128 ( u - 1 ) ^{4}}))

代用を戻す

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

= 32 (\frac{3}{256} ln tan (\frac{x}{2}) + 1 - \frac{3}{256 ( tan ( \frac{x}{2} ) + 1 )} - (- \frac{1}{256 ( tan ( \frac{x}{2} ) + 1 ) ^{2}}) - (- \frac{1}{64 ( tan ( \frac{x}{2} ) + 1 ) ^{3}}) - \frac{1}{128 ( tan ( \frac{x}{2} ) + 1 ) ^{4}} - \frac{3}{256} ln tan (\frac{x}{2}) - 1 - \frac{3}{256 ( tan ( \frac{x}{2} ) - 1 )} - \frac{1}{256 ( tan ( \frac{x}{2} ) - 1 ) ^{2}} - (- \frac{1}{64 ( tan ( \frac{x}{2} ) - 1 ) ^{3}}) - (- \frac{1}{128 ( tan ( \frac{x}{2} ) - 1 ) ^{4}}))

簡素化

= 32 (\frac{3}{256} ln tan (\frac{x}{2}) + 1 - \frac{3}{256 ( tan ( \frac{x}{2} ) + 1 )} + \frac{1}{256 ( tan ( \frac{x}{2} ) + 1 ) ^{2}} + \frac{1}{64 ( tan ( \frac{x}{2} ) + 1 ) ^{3}} - \frac{1}{128 ( tan ( \frac{x}{2} ) + 1 ) ^{4}} - \frac{3}{256} ln tan (\frac{x}{2}) - 1 - \frac{3}{256 ( tan ( \frac{x}{2} ) - 1 )} - \frac{1}{256 ( tan ( \frac{x}{2} ) - 1 ) ^{2}} + \frac{1}{64 ( tan ( \frac{x}{2} ) - 1 ) ^{3}} + \frac{1}{128 ( tan ( \frac{x}{2} ) - 1 ) ^{4}})

定数を解答に追加する

= 32 (\frac{3}{256} ln tan (\frac{x}{2}) + 1 - \frac{3}{256 ( tan ( \frac{x}{2} ) + 1 )} + \frac{1}{256 ( tan ( \frac{x}{2} ) + 1 ) ^{2}} + \frac{1}{64 ( tan ( \frac{x}{2} ) + 1 ) ^{3}} - \frac{1}{128 ( tan ( \frac{x}{2} ) + 1 ) ^{4}} - \frac{3}{256} ln tan (\frac{x}{2}) - 1 - \frac{3}{256 ( tan ( \frac{x}{2} ) - 1 )} - \frac{1}{256 ( tan ( \frac{x}{2} ) - 1 ) ^{2}} + \frac{1}{64 ( tan ( \frac{x}{2} ) - 1 ) ^{3}} + \frac{1}{128 ( tan ( \frac{x}{2} ) - 1 ) ^{4}}) + C

グラフ

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