∫_0^1▒〖4〖sin〗^2 〗 x dx. Created by Grace. science-mathematics-en - mathematics-en

how do you solve this ∫_0^1▒〖4〖sin〗^2 〗 x dx?

May 2021 2 33 Report

how do you solve this ∫_0^1▒〖4〖sin〗^2 〗 x dx?

∫_0^1▒〖4〖sin〗^2 〗 x dx

That is creative typography. :) I'm assuming that the integral to solve is

integral(4*sin^2(x) dx) (from x = 0 to x = 1).

First note that cos(2x) = cos^2(x) - sin^2(x) = 1 - 2sin^2(x), and so

sin^2(x) = (1/2)*(1 - cos(2x)). So now the integral becomes

integral(2*(1 - cos(2x)) dx) (from x=0 to x=1). Now let u = 2x, then du = 2 dx and

integral((1 - cos(u)) du) = u - sin(u) = (2x - sin(2x)) (from x=0 to x=1)

= 2 - sin(2) - (0 - sin(0)) = 2 - sin(2) = 1.0907 to 4 decimal places.

It doesnt make sense/....

It is definite integralâ«4sin²x dx or â«4*x*sin²x dx from 0 to 1???

See my pic for both questions