integral of (sec(5x)+csc(5x))^2

解答

\int (sec (5 x) + csc (5 x))^{2} d x

\frac{1}{5} (tan (5 x) + ln (tan^{2} (5 x)) - cot (5 x)) + C

求解步骤

\int (sec (5 x) + csc (5 x))^{2} d x

乘开

(sec (5 x) + csc (5 x))^{2} : sec^{2} (5 x) + 2 sec (5 x) csc (5 x) + csc^{2} (5 x)

= \int sec^{2} (5 x) + 2 sec (5 x) csc (5 x) + csc^{2} (5 x) d x

使用换元积分法

= \int (sec^{2} (u) + 2 sec (u) csc (u) + csc^{2} (u)) \frac{1}{5} d u

提出常数:

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

= \frac{1}{5} \cdot \int sec^{2} (u) + 2 sec (u) csc (u) + csc^{2} (u) d u

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}{5} (\int sec^{2} (u) d u + \int 2 sec (u) csc (u) d u + \int csc^{2} (u) d u)

\int sec^{2} (u) d u = tan (u)

\int 2 sec (u) csc (u) d u = ln (tan^{2} (u))

\int csc^{2} (u) d u = - cot (u)

= \frac{1}{5} (tan (u) + ln (tan^{2} (u)) - cot (u))

u = 5 x

代回

= \frac{1}{5} (tan (5 x) + ln (tan^{2} (5 x)) - cot (5 x))

解答补常数

= \frac{1}{5} (tan (5 x) + ln (tan^{2} (5 x)) - cot (5 x)) + C

n = 8 \sum \infty cos (\frac{1}{n ^{3}})

d er i v a t i v e f (x) = sin (3 ln (x))

d er i v a t i v e y = 11 x + 11 x + 11 x

\int x^{2} \cdot cos (x) d x

\int \frac{3 x + 4}{( x ^{2} + 4 ) ( 3 - x )} d x