- #1
songoku
- 2,294
- 325
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