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

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Answer 1 · 2018-05-04T09:51:04

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

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#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 · 2018-05-04T09:53:05

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

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 · 2018-05-04T09:59:50

#(5π^2)/72+1-sqrt3/2#

#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#