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受欢迎的微积分 >

integral of cot^5(2bx)

解答

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

解答

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

求解步骤

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

使用换元积分法

= \int \frac{cot ^{5} ( u )}{2 b} d u

提出常数:

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

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

使用换元积分法

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

提出常数:

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

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

长除

\frac{v ^{5}}{1 + v ^{2}} : v^{3} - v + \frac{v}{v ^{2} + 1}

= \frac{1}{2 b} (- \int v^{3} - v + \frac{v}{v ^{2} + 1} 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 b} (- (\int v^{3} d v - \int v d v + \int \frac{v}{v ^{2} + 1} d v))

\int v^{3} d v = \frac{v ^{4}}{4}

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

\int \frac{v}{v ^{2} + 1} d v = \frac{1}{2} ln v^{2} + 1

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

代回

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

化简

\frac{1}{2 b} (- (\frac{cot ^{4} ( 2 b x )}{4} - \frac{cot ^{2} ( 2 b x )}{2} + \frac{1}{2} ln cot^{2} (2 b x) + 1)) : - \frac{1}{2 b} (- \frac{1}{2} cot^{2} (2 b x) + \frac{1}{4} cot^{4} (2 b x) + \frac{1}{2} ln (csc^{2} (2 b x)))

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

解答补常数

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

作图

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