integral of cot^3(10x)

解答

\int cot^{3} (10 x) d x

\frac{1}{10} (- \frac{cot ^{2} ( 10 x )}{2} - ln ∣ sin (10 x) ∣) + C

求解步骤

\int cot^{3} (10 x) d x

使用换元积分法

= \int cot^{3} (u) \frac{1}{10} d u

提出常数:

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

= \frac{1}{10} \cdot \int cot^{3} (u) d u

使用降次积分法

= \frac{1}{10} (- \frac{cot ^{2} ( u )}{2} - \int cot (u) d u)

\int cot (u) d u = ln ∣ sin (u) ∣

= \frac{1}{10} (- \frac{cot ^{2} ( u )}{2} - ln ∣ sin (u) ∣)

u = 10 x

代回

= \frac{1}{10} (- \frac{cot ^{2} ( 10 x )}{2} - ln ∣ sin (10 x) ∣)

解答补常数

= \frac{1}{10} (- \frac{cot ^{2} ( 10 x )}{2} - ln ∣ sin (10 x) ∣) + C

2 \frac{d ^{2} y}{d x ^{2}} + 2 \frac{d y}{d x} + y = 1

\frac{d}{d x} (cos (x^{3} + 2 x))

d er i v a t i v e 3 x^{\frac{4}{3}} - 2 x^{\frac{1}{7}}

x \to - 3 lim (\frac{( x ^{2} + 7 x + 12 )}{x ^{2} - 9})

x \to 3 lim (\frac{x ^{2} - 9}{x + 1})