integral of cot^3(4x)

解答

\int cot^{3} (4 x) d x

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

求解步骤

\int cot^{3} (4 x) d x

使用换元积分法

= \int cot^{3} (u) \frac{1}{4} d u

提出常数:

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

= \frac{1}{4} \cdot \int cot^{3} (u) d u

使用降次积分法

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

\int cot (u) d u = ln ∣ sin (u) ∣

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

u = 4 x

代回

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

解答补常数

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

t an g e n t f (x) = \frac{4 x}{x + 2}, at x = 6

\frac{d}{d x} (3 x + 8)

\int cos^{2} (x t a) n^{3} x d x

d er i v a t i v e 3 u + 5 u

y^{^{'}} = 4 - 9^{2} - 6 x^{5}