Integrating: Solving the Problem of Ln(u) in Answer

  • Thread starter Thread starter crm08
  • Start date Start date
  • Tags Tags
    Integration
crm08
Messages
28
Reaction score
0

Homework Statement



\int(\frac{x}{\sqrt{1-x^{2}}})dx

Homework Equations





The Attempt at a Solution



My calculator tells me that the answer should be -sqrt(1-x^2) but if I pick u = sqrt(1-x^2), then dx = (sqrt(1-x^2)*du)/x, which leaves me with -integral((sqrt(1-x^2)/u)du), the problem I am having is getting rid of the "ln(u)" in my final answer, any suggestions?
 
Physics news on Phys.org
If u^2=1-x^2
2u du =-2x dx => - u du = x dx

Now you'd just get

\frac{-u}{u} du
 
ok got it, thank you
 
rock.freak667 said:
If u^2=1-x^2
2u du =-2x dx => - u du = x dx

Now you'd just get

\frac{-u}{u} du
Another substitution that works is u = 1 - x^2, du = -2xdx.
The integrand then becomes -(1/2)du/u^(1/2), which is also an easy one to integrate.
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...

Similar threads

Back
Top