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受欢迎的微积分 >

integral of (tan(ax)-cot(ax))^3

解答

\int (tan (a x) - cot (a x))^{3} d x

解答

- \frac{4}{a} (- \frac{1}{2} cot^{2} (2 a x) - ln ∣ sin (2 a x) ∣) + C

求解步骤

\int (tan (a x) - cot (a x))^{3} d x

化简

(tan (a x) - cot (a x))^{3} : (tan (a x) - cot (a x))^{2} (tan (a x) - cot (a x))

= \int (tan (a x) - cot (a x))^{2} (tan (a x) - cot (a x)) d x

使用换元积分法

= \int \frac{( tan ( u ) - cot ( u ) ) ^{3}}{a} d u

提出常数:

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

= \frac{1}{a} \cdot \int (tan (u) - cot (u))^{3} d u

化简

(tan (u) - cot (u))^{3} : - 8 cot^{3} (2 u)

= \frac{1}{a} \cdot \int - 8 cot^{3} (2 u) d u

提出常数:

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

= \frac{1}{a} (- 8 \cdot \int cot^{3} (2 u) d u)

使用换元积分法

= \frac{1}{a} (- 8 \cdot \int cot^{3} (v) \frac{1}{2} d v)

提出常数:

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

= \frac{1}{a} (- 8 \cdot \frac{1}{2} \cdot \int cot^{3} (v) d v)

使用降次积分法

= \frac{1}{a} (- 8 \cdot \frac{1}{2} (- \frac{cot ^{2} ( v )}{2} - \int cot (v) d v))

使用三角恒等式改写

= \frac{1}{a} (- 8 \cdot \frac{1}{2} (- \frac{cot ^{2} ( v )}{2} - \int \frac{cos ( v )}{sin ( v )} d v))

使用换元积分法

= \frac{1}{a} (- 8 \cdot \frac{1}{2} (- \frac{cot ^{2} ( v )}{2} - \int \frac{1}{w} d w))

使用积分常见定则:

\int \frac{1}{w} d w = ln (∣ w ∣)

= \frac{1}{a} (- 8 \cdot \frac{1}{2} (- \frac{cot ^{2} ( v )}{2} - ln ∣ w ∣))

代回

= \frac{1}{a} (- 8 \cdot \frac{1}{2} (- \frac{cot ^{2} ( 2 a x )}{2} - ln ∣ sin (2 a x) ∣))

化简

\frac{1}{a} (- 8 \cdot \frac{1}{2} (- \frac{cot ^{2} ( 2 a x )}{2} - ln ∣ sin (2 a x) ∣)) : - \frac{4}{a} (- \frac{1}{2} cot^{2} (2 a x) - ln ∣ sin (2 a x) ∣)

= - \frac{4}{a} (- \frac{1}{2} cot^{2} (2 a x) - ln ∣ sin (2 a x) ∣)

解答补常数

= - \frac{4}{a} (- \frac{1}{2} cot^{2} (2 a x) - ln ∣ sin (2 a x) ∣) + C

作图

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