# Given the function f(x) with property `xf(x)=sin x` , `int_1^(pi/2) f(x)dx` `< cos 1` ?

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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 first `f(x)` , using the property of the function, indicated by the problem, such that:

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

You should consider the following inequality, such that:

`sin x/x < sin x/1` , if `x in (1,pi/2)`

Integrating both sides, yields:

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

Evaluating `int_1^(pi/2) sin x/1 dx` yields:

`int_1^(pi/2) sin x/1 dx = -cos x|_0^(pi/2)`

Using the fundamental theorem of calculus, yields:

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

`int_1^(pi/2) sin x/1 dx = 0 + cos 1`

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

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

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

**Hence, testing if the inequality `int_1^(pi/2) f(x) dx < cos 1 ` holds, yields that `int_1^(pi/2) sin x/x dx < cos 1` is valid, under the given conditions.**