integral of (tan(3x)+cot(3x))^2

解答

\int (tan (3 x) + cot (3 x))^{2} d x

\frac{1}{3} tan (3 x) - \frac{1}{3} cot (3 x) + C

求解步骤

\int (tan (3 x) + cot (3 x))^{2} d x

乘开

(tan (3 x) + cot (3 x))^{2} : tan^{2} (3 x) + 2 tan (3 x) cot (3 x) + cot^{2} (3 x)

= \int tan^{2} (3 x) + 2 tan (3 x) cot (3 x) + cot^{2} (3 x) d x

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

= \int tan^{2} (3 x) d x + \int 2 tan (3 x) cot (3 x) d x + \int cot^{2} (3 x) d x

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

\int 2 tan (3 x) cot (3 x) d x = 2 x

\int cot^{2} (3 x) d x = - x - \frac{1}{3} cot (3 x)

= - x + \frac{1}{3} tan (3 x) + 2 x - x - \frac{1}{3} cot (3 x)

化简

= \frac{1}{3} tan (3 x) - \frac{1}{3} cot (3 x)

解答补常数

= \frac{1}{3} tan (3 x) - \frac{1}{3} cot (3 x) + C

n = 0 \sum \infty \frac{5 ^{n}}{3 ^{n}}

in v erse l a pl a ce \frac{s}{s - 1}

\int (1 - 8 x) (ln (x) + 1) d x

t an g e n t 6 x^{2} + x y + 6 y^{2} = 13, (1, 1)

y^{^{''}} + 9 y = 5 csc^{2} (3 t)