integral of tan(2xse)c^22x

解

\int tan (2 x se) c^{2} 2 x d x

i c^{2} \frac{1}{4 e ^{2} s ^{2}} Li_{2} (- e^{2 \cdot 2 i es x}) + i c^{2} x^{2} - \frac{c ^{2}}{es} x ln (cos (4 es x) + i sin (4 es x) + 1) + C

解答ステップ

\int tan (2 x se) c^{2} \cdot 2 x d x

定数を除く:

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

= 2 c^{2} \cdot \int tan (2 se x) x d x

u置換積分法を適用する

= 2 c^{2} \cdot \int \frac{u tan ( u )}{4 e ^{2} s ^{2}} d u

定数を除く:

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

= 2 c^{2} \frac{1}{4 e ^{2} s ^{2}} \cdot \int u tan (u) d u

非初等積分を使用する:

\int x tan (x) d x = i \frac{1}{2} Li_{2} (- e^{2 i x}) + i \frac{1}{2} x^{2} - x ln (1 + e^{2 i x})

= 2 c^{2} \frac{1}{4 e ^{2} s ^{2}} (i \frac{1}{2} Li_{2} (- e^{2 i u}) + i \frac{1}{2} u^{2} - u ln (1 + e^{2 i u}))

代用を戻す

u = 2 se x

= 2 c^{2} \frac{1}{4 e ^{2} s ^{2}} (i \frac{1}{2} Li_{2} (- e^{2 i 2 se x}) + i \frac{1}{2} (2 se x)^{2} - 2 se x ln (1 + e^{2 i 2 se x}))

簡素化

2 c^{2} \frac{1}{4 e ^{2} s ^{2}} (i \frac{1}{2} Li_{2} (- e^{2 i 2 se x}) + i \frac{1}{2} (2 se x)^{2} - 2 se x ln (1 + e^{2 i 2 se x})) : i c^{2} \frac{1}{4 e ^{2} s ^{2}} Li_{2} (- e^{2 \cdot 2 i es x}) + i c^{2} x^{2} - \frac{c ^{2}}{es} x ln (cos (4 es x) + i sin (4 es x) + 1)

= i c^{2} \frac{1}{4 e ^{2} s ^{2}} Li_{2} (- e^{2 \cdot 2 i es x}) + i c^{2} x^{2} - \frac{c ^{2}}{es} x ln (cos (4 es x) + i sin (4 es x) + 1)

定数を解答に追加する

= i c^{2} \frac{1}{4 e ^{2} s ^{2}} Li_{2} (- e^{2 \cdot 2 i es x}) + i c^{2} x^{2} - \frac{c ^{2}}{es} x ln (cos (4 es x) + i sin (4 es x) + 1) + C