integral of sin(tan(x))

解答

\int sin (tan (x)) d x

\frac{1}{4 e} ((e^{2} - 1) Ci (i - tan (x)) + (e^{2} - 1) Ci (tan (x) + i) + i (1 + e^{2}) (Si (i - tan (x)) + Si (tan (x) + i))) + C

求解步骤

\int sin (tan (x)) d x

使用换元积分法

= \int \frac{sin ( u )}{1 + u ^{2}} d u

这是非初等函数积分:

\int \frac{sin ( x )}{1 + x ^{2}} d x = \frac{1}{4 e} ((e^{2} - 1) Ci (i - x) + (e^{2} - 1) Ci (x + i) + i (1 + e^{2}) (Si (i - x) + Si (x + i)))

= \frac{1}{4 e} ((e^{2} - 1) Ci (i - u) + (e^{2} - 1) Ci (u + i) + i (1 + e^{2}) (Si (i - u) + Si (u + i)))

u = tan (x)

代回

= \frac{1}{4 e} ((e^{2} - 1) Ci (i - tan (x)) + (e^{2} - 1) Ci (tan (x) + i) + i (1 + e^{2}) (Si (i - tan (x)) + Si (tan (x) + i)))

解答补常数

= \frac{1}{4 e} ((e^{2} - 1) Ci (i - tan (x)) + (e^{2} - 1) Ci (tan (x) + i) + i (1 + e^{2}) (Si (i - tan (x)) + Si (tan (x) + i))) + C

\int \frac{x + 1}{x - 1} d x

\int e^{e^{e^{x}}} d e^{e^{x}}

\int \frac{1}{2 + cos ^{2} ( x )} d x

\int 2 sin (x) - 2 x^{\frac{5}{2}} + C d x

\int (\frac{6}{5 + 5 x ^{2}}) d x