Use the properties of integrals to verify the inequality

[mgaddafi86]

Homework Statement

∫(from pi/4 to pi/2)sin x/x ≤ 1/√2.

Homework Equations

The Attempt at a Solution

I know the pi/4≤x≤pi/2 and so 1/√2 ≤ sin x ≤ 1 and i have tried to manipulate this to no end and it has annoyed the living daylights out of me

 

[Mark44]

Homework Statement

∫(from pi/4 to pi/2)sin x/x ≤ 1/√2.

Homework Equations

The Attempt at a Solution

I know the pi/4≤x≤pi/2 and so 1/√2 ≤ sin x ≤ 1 and i have tried to manipulate this to no end and it has annoyed the living daylights out of me

sin(x) is increasing on your interval, and so is x. What about sin(x)/x? Is this function increasing, decreasing, or neither? Can you find upper and lower bounds on the values of sin(x)/x on the given interval?

 

[mgaddafi86]

Yes I tried that,
the derivative of the function sin(x)/x would be: [x*cos(x)-sin(x)]/[x^2]

the denominator will be bigger than 0. sin(x) is between 1/sqrt(2) and 1 cos(x) is between 0 and 1/sqrt(2). If I was sure that x is smaller that on or equal to one then I could conclude that the derivative is negative because sinx >= cos x on the interval. But x is between pi/4(smaller than one) and pi/2(bigger than one)

 

[mgaddafi86]

YES NOW I GOT IT!

[x*cos(x)-sin(x)]/[x^2] will be bigger than 0 because:

if 0<=x<=pi/2 then x<=tan(x) so x<=sin(x)/cos(x) xcos(x)<=sin(x) xcos(x)-sin(x)<=0

thus the derivative is negative or 0 on the interval
so sin(x)/x <= sin(pi/4)/(pi/4) because it is decreasing

so ∫(from pi/4 to pi/2)sin x/x <= ∫(from pi/4 to pi/2)1/sqrt(2)*4/pi = 1/sqrt(2)*4/pi*(pi/2-pi/4)= 1/sqrt(2)
YEEEEEEEEEEEEE. I have been working on this stupid problem for 3 days.

Thank you for helping out.