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

Question

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

Samantha Adkison · Accepted Answer

#(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#

Paisley Johnson · Answer

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

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#

Isaiah Allen · Answer

#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#

Emma Aldridge · Answer

undefined To integrate ∫ (x + cos(x)) dx from π/3 to π/2, you first integrate each term separately, then evaluate the integral at the upper and lower bounds and find the difference. The integral of x with respect to x is x^2/2, and the integral of cos(x) with respect to x is sin(x). Then apply the Fundamental Theorem of Calculus to find the definite integral. Finally, substitute the upper bound (π/2) and subtract the result when substituting the lower bound (π/3).