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

1 Answer
Mar 7, 2018

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

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