Show that $\int_{0}^{1} \frac{sin x}{\sqrt{x^{2} + 1}} d x < \sqrt{2} - 1$ $int_0^1sinx/sqrt(x^2+1)dx<sqrt2-1$ ?

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Show that $\int_{0}^{1} \frac{sin x}{\sqrt{x^{2} + 1}} d x < \sqrt{2} - 1$ $int_0^1sinx/sqrt(x^2+1)dx<sqrt2-1$ ?

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Answer 1 · 2018-03-07T14:51:33

See explanation

Explanation:

We want to show

$\int_{0}^{1} \frac{sin (x)}{\sqrt{x^{2} + 1}} d x < \sqrt{2} - 1$ $int_0^1sin(x)/sqrt(x^2+1)dx < sqrt(2)-1$

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

$\int_{0}^{1} \frac{x}{\sqrt{x^{2} + 1}} d x < \sqrt{2} - 1$ $int_0^1x/sqrt(x^2+1)dx < sqrt(2)-1$

Then our first statement must also be true

The new integral is a simple substitution problem

$\int_{0}^{1} \frac{x}{\sqrt{x^{2} + 1}} = {[\sqrt{x^{2} + 1}]}_{0}^{1} = \sqrt{2} - 1$ $int_0^1x/sqrt(x^2+1)=[sqrt(x^2+1)]_0^1=sqrt(2)-1$

The last step is to notice that $sin (x) = x \Rightarrow x = 0$ $sin(x)=x=>x=0$

Therefore we can conclude

$\int_{0}^{1} \frac{sin (x)}{\sqrt{x^{2} + 1}} d x < \sqrt{2} - 1$ $int_0^1sin(x)/sqrt(x^2+1)dx < sqrt(2)-1$

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Show that $\int_{0}^{1} \frac{sin x}{\sqrt{x^{2} + 1}} d x < \sqrt{2} - 1$ $int_0^1sinx/sqrt(x^2+1)dx<sqrt2-1$ ?

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Explanation:

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Show that ∫10sinx√x2+1dx<√2−1int_0^1sinx/sqrt(x^2+1)dx<sqrt2-1 ?

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Show that $\int_{0}^{1} \frac{sin x}{\sqrt{x^{2} + 1}} d x < \sqrt{2} - 1$ $int_0^1sinx/sqrt(x^2+1)dx<sqrt2-1$ ?