Can substitution help solve this improper integral?

  • Thread starter Thread starter regnar
  • Start date Start date
  • Tags Tags
    Integrals
regnar
Messages
24
Reaction score
0
I've been stuck on this problem for quite some time and I get as far as setting up the problem. They are asking me to evaluate this integral:

\int_0^1{\frac{(6ln4x)}{\sqrt{x}}dx and I get as far as:

\lim_{b \to 0^+}{6\int_b^1{\frac{(ln4x)}{\sqrt{x}}
 
Physics news on Phys.org
Substituting x = t^2 seems to work...
 
I'm not sure what you mean or where t^2 came from.
 
It's called substitution.

The substitution is t=\sqrt{x} or t^2 = x
 
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