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受欢迎的微积分 >

integral of csc^3(2x+1)

解答

\int csc^{3} (2 x + 1) d x

解答

\frac{1}{16} (- cot^{2} (\frac{2 x + 1}{2}) + 4 ln tan (\frac{2 x + 1}{2}) + tan^{2} (\frac{2 x + 1}{2})) + C

求解步骤

\int csc^{3} (2 x + 1) d x

使用三角恒等式改写

= \int \frac{1}{sin ^{3} ( 2 x + 1 )} d x

使用换元积分法

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

提出常数:

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

= \frac{1}{2} \cdot \int \frac{1}{sin ^{3} ( u )} d u

使用换元积分法

= \frac{1}{2} \cdot \int \frac{( 1 + v ^{2} ) ^{2}}{4 v ^{3}} d v

提出常数:

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

= \frac{1}{2} \cdot \frac{1}{4} \cdot \int \frac{( 1 + v ^{2} ) ^{2}}{v ^{3}} d v

乘开

\frac{( 1 + v ^{2} ) ^{2}}{v ^{3}} : \frac{1}{v ^{3}} + \frac{2}{v} + v

= \frac{1}{2} \cdot \frac{1}{4} \cdot \int \frac{1}{v ^{3}} + \frac{2}{v} + v d v

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

= \frac{1}{2} \cdot \frac{1}{4} (\int \frac{1}{v ^{3}} d v + \int \frac{2}{v} d v + \int v d v)

\int \frac{1}{v ^{3}} d v = - \frac{1}{2 v ^{2}}

\int \frac{2}{v} d v = 2 ln ∣ v ∣

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

= \frac{1}{2} \cdot \frac{1}{4} (- \frac{1}{2 v ^{2}} + 2 ln ∣ v ∣ + \frac{v ^{2}}{2})

代回

= \frac{1}{2} \cdot \frac{1}{4} (- \frac{1}{2 tan ^{2} ( \frac{2 x + 1}{2} )} + 2 ln tan (\frac{2 x + 1}{2}) + \frac{tan ^{2} ( \frac{2 x + 1}{2} )}{2})

化简

\frac{1}{2} \cdot \frac{1}{4} (- \frac{1}{2 tan ^{2} ( \frac{2 x + 1}{2} )} + 2 ln tan (\frac{2 x + 1}{2}) + \frac{tan ^{2} ( \frac{2 x + 1}{2} )}{2}) : \frac{1}{16} (- cot^{2} (\frac{2 x + 1}{2}) + 4 ln tan (\frac{2 x + 1}{2}) + tan^{2} (\frac{2 x + 1}{2}))

= \frac{1}{16} (- cot^{2} (\frac{2 x + 1}{2}) + 4 ln tan (\frac{2 x + 1}{2}) + tan^{2} (\frac{2 x + 1}{2}))

解答补常数

= \frac{1}{16} (- cot^{2} (\frac{2 x + 1}{2}) + 4 ln tan (\frac{2 x + 1}{2}) + tan^{2} (\frac{2 x + 1}{2})) + C

作图

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