- #1

- 1,815

- 177

## Homework Statement

Find using substitution [tex]x=\sin \theta[/tex]

[tex]\int _0^{0.5}\frac{x}{\sqrt{1-x^2}}dx[/tex]

## Homework Equations

integration

## The Attempt at a Solution

[tex]\int _0^{0.5}\frac{x}{\sqrt{1-x^2}}dx[/tex]

[tex]=\int_0^{\frac{\pi}{6}}\sin \theta\;d\theta[/tex]

[tex]=1-\frac{1}{2}\sqrt{3}[/tex]

What I want to ask is about changing the upper bound of the integral.

For x = 0.5 :

[tex]0.5 = \sin \theta[/tex]

Here I choose [tex]\theta = \frac{\pi}{6}[/tex]. But what if I choose [tex]\theta = \frac{5}{6}\pi\;??[/tex]

So, the integral :

[tex]=\int_0^{\frac{5}{6}\pi}\sin \theta\;d\theta[/tex]

[tex]=1+\frac{1}{2}\sqrt{3}[/tex]

Thanks