Question

How to solve `int_(-pi/3)^(pi/3)` `(|x| + tan x)^2dx`?

`x` is an absolute value

Answer 1

`int_(-pi/3)^(pi/3) (absx+tanx)^2dx = (2pi^3)/81-pi/3 + sqrt3`

Explanation:

Using the additivity of the integral:

`int_(-pi/3)^(pi/3) (absx+tanx)^2dx = int_(-pi/3)^0 (absx+tanx)^2dx + int_0^(pi/3) (absx+tanx)^2dx`

In the first integral substitute `t=-x`:

`int_(-pi/3)^0 (absx+tanx)^2dx = int_(pi/3)^0 (abs(-t)+tan(-t))^2d(-t)`

`int_(-pi/3)^0 (absx+tanx)^2dx = int_0^(pi/3) (abs(t)-tant)^2dt`

Change the name of the integration variable to `x` again for clarity:

`int_(-pi/3)^(pi/3) (absx+tanx)^2dx = int_0^(pi/3) (absx+tanx)^2dx + int_0^(pi/3) (abs(x)-tanx)^2dx`

as in the interval of integration `x` is positive, then `absx = x`:

`int_(-pi/3)^(pi/3) (absx+tanx)^2dx = int_0^(pi/3) (x+tanx)^2dx + int_0^(pi/3) (x-tanx)^2dx`

using the linearity of the integral:

`int_(-pi/3)^(pi/3) (absx+tanx)^2dx = int_0^(pi/3) ((x+tanx)^2+(x-tanx)^2)dx`

`int_(-pi/3)^(pi/3) (absx+tanx)^2dx = int_0^(pi/3) (x^2+2xtanx+tan^2x+x^2-2xtanx+tan^2x)dx`

`int_(-pi/3)^(pi/3) (absx+tanx)^2dx = 2int_0^(pi/3) (x^2+tan^2x)dx`

`int_(-pi/3)^(pi/3) (absx+tanx)^2dx = 2int_0^(pi/3) x^2dx+2int_0^(pi/3) tan^2xdx`

Using the trigonometric identity:

`tan^2x=sec^2x-1`

`int_(-pi/3)^(pi/3) (absx+tanx)^2dx = 2int_0^(pi/3) x^2dx+2int_0^(pi/3) (sec^2x-1)dx`

`int_(-pi/3)^(pi/3) (absx+tanx)^2dx = 2int_0^(pi/3) x^2dx+2int_0^(pi/3) sec^2xdx - 2int_0^(pi/3)dx`

`int_(-pi/3)^(pi/3) (absx+tanx)^2dx = 2[x^3/3+tanx-x]_0^(pi/3)`

`int_(-pi/3)^(pi/3) (absx+tanx)^2dx = (2pi^3)/81-pi/3 + sqrt3`