# How do you evaluate the integral 1/(sqrt(49-x^2)) from 0 to 7sqrt(3/2)?

Feb 19, 2015

Hello,

The result is $\arcsin \left(\sqrt{\frac{3}{2}}\right)$, but cf. the remark at the end of my text.

Write $\setminus \frac{1}{\setminus \sqrt{49 - {x}^{2}}} = \setminus \frac{1}{\setminus \sqrt{49 \left(1 - {\left(\frac{x}{7}\right)}^{2}\right)}} = \setminus \frac{1}{7 \setminus \sqrt{1 - {\left(\frac{x}{7}\right)}^{2}}}$

Change the variable in the integral :
Take $u = \frac{x}{7}$, therefore :
- $\mathrm{du} = \setminus \frac{d x}{7}$ or $\mathrm{dx} = 7 \mathrm{du}$,
- if $x = 0$, then $u = 0$,
- if $x = 7 \setminus \sqrt{\frac{3}{2}}$, then $u = \sqrt{\frac{3}{2}}$.

So, $\setminus {\int}_{0}^{7 \sqrt{\frac{3}{2}}} \setminus \frac{\mathrm{dx}}{\sqrt{49 - {x}^{2}}} = \setminus {\int}_{0}^{\sqrt{\frac{3}{2}}} \setminus \frac{7 \mathrm{du}}{7 \setminus \sqrt{1 - {u}^{2}}} = \setminus {\int}_{0}^{\sqrt{\frac{3}{2}}} \setminus \frac{\mathrm{du}}{\sqrt{1 - {u}^{2}}}$

You recognize the derivative of $\arcsin$, so

$\setminus {\int}_{0}^{7 \sqrt{\frac{3}{2}}} \setminus \frac{\mathrm{dx}}{\sqrt{49 - {x}^{2}}} = {\left[\arcsin \left(u\right)\right]}_{0}^{\sqrt{\frac{3}{2}}} = \arcsin \left(\sqrt{\frac{3}{2}}\right)$.

Remark. If it's $7 \frac{\sqrt{3}}{2}$ and not $7 \sqrt{\frac{3}{2}}$, the result is easier, because $\arcsin \left(\frac{\sqrt{3}}{2}\right) = \frac{\pi}{3}$.

Feb 19, 2015

The answer is: $\frac{\pi}{3}$.

I think there is a little mistake in your writing, are you sure that the limits of integration weren't $0$ and $7 \frac{\sqrt{3}}{2}$?

${\int}_{0}^{7 \frac{\sqrt{3}}{2}} \frac{1}{\sqrt{49 - {x}^{2}}} \mathrm{dx} = {\int}_{0}^{7 \frac{\sqrt{3}}{2}} \frac{1}{\sqrt{49 \left(1 - {x}^{2} / 49\right)}} \mathrm{dx} =$

${\int}_{0}^{7 \frac{\sqrt{3}}{2}} \frac{1}{7 \sqrt{1 - {\left(\frac{x}{7}\right)}^{2}}} \mathrm{dx} = {\int}_{0}^{7 \frac{\sqrt{3}}{2}} \frac{\frac{1}{7}}{\sqrt{1 - {\left(\frac{x}{7}\right)}^{2}}} \mathrm{dx} =$

$= {\left[\arcsin \left(\frac{x}{7}\right)\right]}_{0}^{7 \frac{\sqrt{3}}{2}} = \arcsin \left(\frac{\sqrt{3}}{2}\right) - \arcsin 0 = \frac{\pi}{3}$.

This is remembering:

$\int \frac{f ' \left(x\right)}{\sqrt{1 - {\left[f \left(x\right)\right]}^{2}}} \mathrm{dx} = \arcsin f \left(x\right) + c .$