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

Mar 7, 2018

See explanation

#### Explanation:

We want to show

${\int}_{0}^{1} \sin \frac{x}{\sqrt{{x}^{2} + 1}} \mathrm{dx} < \sqrt{2} - 1$

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 $\textcolor{red}{\sin \left(x\right) \le x}$
Thus, the value of the integrand will also be bigger, for all positive real numbers, if we substitute $x = \sin \left(x\right)$, so if we can show

${\int}_{0}^{1} \frac{x}{\sqrt{{x}^{2} + 1}} \mathrm{dx} < \sqrt{2} - 1$

Then our first statement must also be true

The new integral is a simple substitution problem

${\int}_{0}^{1} \frac{x}{\sqrt{{x}^{2} + 1}} = {\left[\sqrt{{x}^{2} + 1}\right]}_{0}^{1} = \sqrt{2} - 1$

The last step is to notice that $\sin \left(x\right) = x \implies x = 0$

Therefore we can conclude

${\int}_{0}^{1} \sin \frac{x}{\sqrt{{x}^{2} + 1}} \mathrm{dx} < \sqrt{2} - 1$