What is the integral of xtan(2x) ?

The integral of xtan(2x) is 1/4 (i 1/2 \Li_{2}(-e^{2*2ix})+2ix^2-2xln(1+cos(4x)+isin(4x)))+C

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integral of xtan(2x)

\int x tan (2 x) d x

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

Solution steps

\int x tan (2 x) d x

Apply u-substitution

: \int \frac{u tan ( u )}{4} d u

= \int \frac{u tan ( u )}{4} d u

Take the constant out:

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

= \frac{1}{4} \cdot \int u tan (u) d u

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})

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

Substitute back

u = 2 x

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

Simplify

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

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

Add a constant to the solution

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

\int (cos (7 x)) d x

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

t \cdot e^{- \frac{t}{2}}

3 x^{2} - 6 x + 4

\frac{d}{d t} (t e^{a t})

What is the integral of xtan(2x) ?
The integral of xtan(2x) is 1/4 (i 1/2 \Li_{2}(-e^{2*2ix})+2ix^2-2xln(1+cos(4x)+isin(4x)))+C