Question

How do you integrate `int x+cosx` from [pi/3, pi/2]?

Answer 1

The answer `int _(pi/3)^(pi/2)x+cosx*dx=0.8193637907356557`

Explanation:

show below

`int _(pi/3)^(pi/2)x+cosx*dx=[1/2x^2+sinx]_(pi/3)^(pi/2)`

`[pi^2/8+sin(pi/2)]-[pi^2/18+sin(pi/3)]=(5*pi^2-4*3^(5/2)+72)/72=0.8193637907356557`

Answer 2

`int_(pi/3)^(pi/2) (x+cosx)dx =1+(5pi^2-36sqrt3)/72`

Explanation:

Using the linearity of the integral:

`int_(pi/3)^(pi/2) (x+cosx)dx = int_(pi/3)^(pi/2)xdx + int_(pi/3)^(pi/2) cosxdx`

Now:

`int_(pi/3)^(pi/2)xdx = [x^2/2]_(pi/3)^(pi/2) = pi^2/8-pi^2/18 = (5pi^2)/72`

`int_(pi/3)^(pi/2) cosxdx = [sinx]_(pi/3)^(pi/2) = sin(pi/2)-sin(pi/3) = 1-sqrt3/2`

Then:

`int_(pi/3)^(pi/2) (x+cosx)dx =1+(5pi^2-36sqrt3)/72`

Answer 3

`(5π^2)/72+1-sqrt3/2`

Explanation:

`int_(π/3)^(π/2)(x+cosx)dx` `=`

`int_(π/3)^(π/2)xdx+int_(π/3)^(π/2)cosxdx` `=`

`[x^2/2]_(π/3)^(π/2)` `+` `[sinx]_(pi/3)^(π/2)` `=`

`(π^2/4)/2-(π^2/9)/2+sin(π/2)-sin(π/3)` `=`

`π^2/8-π^2/18+1-sqrt3/2` `=`

`(5π^2)/72+1-sqrt3/2`