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受欢迎的微积分 >

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

解答

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

解答

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

求解步骤

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

使用三角恒等式改写

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

使用换元积分法

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

提出常数:

\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

将

\frac{u ^{4}}{( 1 - u ^{2} ) ^{5}}

用部份分式展开:

\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

使用积分加法定则:

\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}}))

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

代回

= 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}}))

化简

= 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}})

解答补常数

= 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

作图

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