# Explain please how is demonstrate definite integral from 1 to pie/2 f(x) < cos 1? xf(x)= sinx

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### 1 Answer

You need to prove that `int_1^(pi/2) f(x)dx < cos 1` , hence, you need to find `f(x)` , using the information provided by the problem, such that:

`x*f(x) = sin x => f(x) = (sin x)/x`

Replacing `(sin x)/x` for `f(x)` yields:

`int_1^(pi/2) (sin x)/x dx < cos 1`

Since `x in [1,pi/2]` , hence, the following inequality holds:

`(sin x)/x < (sin x)/1`

Integrating both sides, yields:

`int_1^(pi/2) (sin x)/x dx < int_1^(pi/2) (sin x)/1 dx`

Evaluating the definite integral yields:

`int_1^(pi/2) (sin x)/1 dx = -cos x|_1^(pi/2)`

Using the fundamental theorem of calculus yields:

`int_1^(pi/2) (sin x)/1 dx = -cos (pi/2) - (- cos 1)`

Since `cos (pi/2) = 0` yields:

`int_1^(pi/2) (sin x)/1 dx = cos 1`

Replacing `cos 1` for `int_1^(pi/2) (sin x)/1 dx` in inequality considered yields:

`int_1^(pi/2) (sin x)/x dx < int_1^(pi/2) (sin x)/1 dx = cos 1`

**Hence, testing if the inequality` int_1^(pi/2) (sin x)/x dx < cos 1` holds, using the inequality `(sin x)/x < (sin x)/1` , `x in [1,pi/2]` yields `int_1^(pi/2) (sin x)/x dx < int_1^(pi/2) (sin x)/1 dx = cos 1` valid.**