Use the properties of integrals to verify the inequality without evaluating the integrals.

(sqrt(2)pi)/24 is greater than or equal to integrate from pi/6 to pi/4 of cos(x)dx is less than or equal to (sqrt(3)pi)/24

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You should remember that if `a =< x =< b` , such that`m =< f(x) =< M` , then `m(b-a) =< int_a^b f(x) dx =< M(b-a).`

You need to prove that `pi*sqrt2/24 =< int_(pi/6)^(pi/4) cos x dx =< pi*sqrt3/24` , hence, using the property above yields:

`m(pi/4 - pi/6) =< int_(pi/6)^(pi/4) cos x dx =< M(pi/4 - pi/6)`

`m*(pi/12) =< int_(pi/6)^(pi/4) cos x dx =< M(pi/12) `

Notice that `pi*sqrt2/24 = (pi/12)*(sqrt2/2)=(pi/12)*(cos(pi/4))` and `pi*sqrt3/24 = (pi/12)*(sqrt3/2)= (pi/12)*(cos(pi/6))` , hence `M = (cos(pi/6))` and `m =cos(pi/4)`

Reasoning by analogy yields:

`(cos(pi/4))*(pi/12) =< int_(pi/6)^(pi/4) cos x dx =< (cos(pi/6))(pi/12) => cos(pi/4) =< cos x =< cos(pi/6)`

Since `pi/4 > pi/6` and `cos(pi/4) =< cos(pi/6), ` then the function `cos x` needs to decrease over `[pi/6;pi/4].`

**Notice that the cosine function decreases over `[0,pi]` , hence the inequality cos(pi/4) =< cos x =< cos(pi/6) holds, then `pi*sqrt2/24 =< int_(pi/6)^(pi/4) cos x dx =< pi*sqrt3/24` .**

**Sources:**

2p/24

Let Int[p/6 to p/4] cosxdx = I

We know (cosx)' = -sinx < 0 in the 1st quadrant. So cosx is a decreasing function in the 1st quadrant.

=>

cos(p/6)(P/3-p/4)>Int[p/6 to p/4] cosxdx = I > cos(p/6)(P/3-p/4)

(sqrt3)p/24 > I > p/24.

=>

**2p/24 < (sqrt3)p/24 > I > p/**24, which proves the inequality.

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