`6int_0^1tan {:x/2:} dx`

Let `u=x/2`, then `du=1/2\ dx`.

`6int_0^1tan {:x/2:}dx =6(-2)[ln|cos {:x/2:}|]_0^1`

`=-12[ln(cos {:1/2:})-ln\ (cos 0)]`

`=1.5670`

Of course, `x` is in radians.

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