How to Solve the Integral of \(\sqrt[3]{1-x^7}-\sqrt[7]{1-x^3}\)?

  • Thread starter Thread starter blockcolder
  • Start date Start date
  • Tags Tags
    Integral
blockcolder
Messages
1
Reaction score
0

Homework Statement



\int_0^1 \sqrt[3]{1-x^7}-\sqrt[7]{1-x^3} dx

Homework Equations



None

The Attempt at a Solution



I tried using the substitutions u=\sqrt[3]{1-x^7} and u=\sqrt[7]{1-x^3} to no avail and I couldn't think of any more substitutions. Any suggestions?
 
Physics news on Phys.org
express the roots as fractional powers
treat the two terms separately

consider the geometry of the integral in terms of the relation
y^3 + x^7 = 1

the roles of x and y swap in the second integral don't they?
 
blockcolder said:

Homework Statement



\int_0^1 \sqrt[3]{1-x^7}-\sqrt[7]{1-x^3} dx

Homework Equations



None

The Attempt at a Solution



I tried using the substitutions u=\sqrt[3]{1-x^7} and u=\sqrt[7]{1-x^3} to no avail and I couldn't think of any more substitutions. Any suggestions?

In the first integral, let y = x^7. Then look up "Beta function".

RGV
 
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