integral of sin(tan(x))

解

\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

解答ステップ

\int sin (tan (x)) d x

u置換積分法を適用する

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

非初等積分を使用する:

\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)))

代用を戻す

u = tan (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)))

定数を解答に追加する

= \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