integral of-xtan(x)

解答

\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 + cos (2 x) + i sin (2 x)) + C

求解步骤

\int - x tan (x) d x

提出常数:

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

= - \int x tan (x) d x

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})

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

化简

- (i \frac{1}{2} Li_{2} (- e^{2 i x}) + i \frac{1}{2} x^{2} - x ln (1 + e^{2 i x})) : - i \frac{1}{2} Li_{2} (- e^{2 i x}) - i \frac{1}{2} x^{2} + x ln (1 + cos (2 x) + i sin (2 x))

= - i \frac{1}{2} Li_{2} (- e^{2 i x}) - i \frac{1}{2} x^{2} + x ln (1 + cos (2 x) + i sin (2 x))

解答补常数

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

x \to 1 lim (2 x^{2} - 5 x - 1)

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