Question

Show that `int_0^1sinx/sqrt(x^2+1)dx<sqrt2-1` ?

Answer 1

See explanation

Explanation:

We want to show

`int_0^1sin(x)/sqrt(x^2+1)dx < sqrt(2)-1`

This is a quite "ugly" integral, so our approach will not be to solve this integral, but compare it to a "nicer" integral

We now that for all positive real numbers `color(red)(sin(x)<=x)`
Thus, the value of the integrand will also be bigger, for all positive real numbers, if we substitute `x=sin(x)`, so if we can show

`int_0^1x/sqrt(x^2+1)dx < sqrt(2)-1`

Then our first statement must also be true

The new integral is a simple substitution problem

`int_0^1x/sqrt(x^2+1)=[sqrt(x^2+1)]_0^1=sqrt(2)-1`

The last step is to notice that `sin(x)=x=>x=0`

Therefore we can conclude

`int_0^1sin(x)/sqrt(x^2+1)dx < sqrt(2)-1`