What is the integral of-xtan(x) ?

The integral of-xtan(x) is -i 1/2 \Li_{2}(-e^{2ix})-i 1/2 x^2+xln(1+cos(2x)+isin(2x))+C

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integral of-xtan(x)

\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

Solution steps

\int - x tan (x) d x

Take the constant out:

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

= - \int x tan (x) d x

Use the nonelementary integral:

\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}))

Simplify

- (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))

Add a constant to the solution

= - 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)

6 e^{- 5 t} + e^{3 t} + 5 t^{3} - 9

\frac{\partial}{\partial u} (u - v)

\int \frac{( 4 - x ^{2} ) ^{\frac{3}{2}}}{x ^{6}} d x

\int \frac{2}{x} y d x

What is the integral of-xtan(x) ?
The integral of-xtan(x) is -i 1/2 \Li_{2}(-e^{2ix})-i 1/2 x^2+xln(1+cos(2x)+isin(2x))+C