`(pisqrt(2))/24 <= int_(pi/6)^(pi/4)cos(x)dx <= (pisqrt(3))/24` Use the properties of integrals to verify the inequality without evaluating the integrals.

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Chapter 5, 5.2 - Problem 58 - Calculus: Early Transcendentals (7th Edition, James Stewart).
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You need to use the mean value thorem to verify the given inequality, such that:

int_a^b f(x)dx = (b-a)*f(c), c in (a,b)

Replacing cos x for f(x) and pi/6 for a, pi/4 for b, yields:

int_(pi/6)^(pi/4) (cos x) dx= (pi/4 - pi/6)f(c) = (pi/12)f(c)

You need to prove that (pi*sqrt2)/24 <= int_(pi/6)^(pi/4) (cos x) dx <= (pi*sqrt3)/24, hence, according to mean value theorem, (pi*sqrt2)/24 <=(-pi/12)f(c)<= (pi*sqrt3)/24.

You need to check the monotony of the function f(x), hence, you need to find the derivative using the chain rule and to verify if it is positive or negative on interval (pi/6,pi/4), such that:

`f(x) = (cos x) => f'(x) = -sin x`

You need to notice that f(x) decreases on `(pi/6,pi/4).`

Hence, for `c in(pi/6,pi/4)` yields:

`pi/6<c<pi/4 => f(pi/6)>f(c)>f(pi/4)`

You need to evalute `f(pi/6)` and `f(pi/4)` , such that:

`f(pi/6) = cos(pi/6) = sqrt3/2`

`f(pi/4) = cos(pi/4) = sqrt2/2`

Hence, replacing the found values in inequality `f(pi/6) >f(c)>f(pi/4) ` , yields:

`sqrt3/2> f(c)>sqrt2/2`

You need to obtain `(pi/12)f(c)` , hence, multiplying by `pi/12` both sides yields:

`(sqrt2/2)*(pi/12) < (pi/12)f(c) < (pi/12)(sqrt3/2)`

`(pi*sqrt2)/24< (pi/12)f(c) <(pi*sqrt3)/24`

But `(pi/12)f(c) =int_(pi/6)^(pi/4) (cos x) dx` and `(pi*sqrt2)/24< (pi/12)f(c) <(pi*sqrt3)/24,` hence `(pi*sqrt2)/24 <= int_(pi/6)^(pi/4) (cos x) dx <= (pi*sqrt3)/24.`

Hence, using mean value theorem, yields that the inequality `(pi*sqrt2)/24 <= int_(pi/6)^(pi/4) (cos x) dx <= (pi*sqrt3)/24` holds. Notice that it is no need to evaluate the integral to prove the inequality.

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