What is the integral of xan(x) ?

The integral of xan(x) is d(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 xan(x)

\int x d tan (x) d x

d (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 d tan (x) d x

Take the constant out:

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

= d \cdot \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})

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

Apply imaginary number rule:

e^{ia} = cos (a) + i sin (a)

= d (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

= d (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

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