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受欢迎的微积分 >

integral of tan(4xse)c^24x

解答

\int tan (4 x se) c^{2} 4 x d x

解答

i c^{2} \frac{1}{8 e ^{2} s ^{2}} Li_{2} (- e^{2 \cdot 4 i es x}) + 2 i c^{2} x^{2} - \frac{c ^{2}}{es} x ln (cos (8 es x) + i sin (8 es x) + 1) + C

求解步骤

\int tan (4 x se) c^{2} \cdot 4 x d x

提出常数:

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

= 4 c^{2} \cdot \int tan (4 se x) x d x

使用换元积分法

= 4 c^{2} \cdot \int \frac{u tan ( u )}{16 e ^{2} s ^{2}} d u

提出常数:

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

= 4 c^{2} \frac{1}{16 e ^{2} s ^{2}} \cdot \int u tan (u) d u

这是非初等函数积分:

\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})

= 4 c^{2} \frac{1}{16 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}))

u = 4 se x

代回

= 4 c^{2} \frac{1}{16 e ^{2} s ^{2}} (i \frac{1}{2} Li_{2} (- e^{2 i 4 se x}) + i \frac{1}{2} (4 se x)^{2} - 4 se x ln (1 + e^{2 i 4 se x}))

化简

4 c^{2} \frac{1}{16 e ^{2} s ^{2}} (i \frac{1}{2} Li_{2} (- e^{2 i 4 se x}) + i \frac{1}{2} (4 se x)^{2} - 4 se x ln (1 + e^{2 i 4 se x})) : i c^{2} \frac{1}{8 e ^{2} s ^{2}} Li_{2} (- e^{2 \cdot 4 i es x}) + 2 i c^{2} x^{2} - \frac{c ^{2}}{es} x ln (cos (8 es x) + i sin (8 es x) + 1)

= i c^{2} \frac{1}{8 e ^{2} s ^{2}} Li_{2} (- e^{2 \cdot 4 i es x}) + 2 i c^{2} x^{2} - \frac{c ^{2}}{es} x ln (cos (8 es x) + i sin (8 es x) + 1)

解答补常数

= i c^{2} \frac{1}{8 e ^{2} s ^{2}} Li_{2} (- e^{2 \cdot 4 i es x}) + 2 i c^{2} x^{2} - \frac{c ^{2}}{es} x ln (cos (8 es x) + i sin (8 es x) + 1) + C

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