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

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SUMMARY

The integral \(\int_0^1 \sqrt[3]{1-x^7}-\sqrt[7]{1-x^3} dx\) can be approached by using substitutions and the properties of the Beta function. A suggested substitution is \(y = x^7\), which simplifies the first term. The discussion emphasizes treating the two terms separately and considering the geometric interpretation of the integral, particularly the relationship \(y^3 + x^7 = 1\).

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Homework Statement



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

Homework Equations



None

The Attempt at a Solution



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

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

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

Homework Statement



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

Homework Equations



None

The Attempt at a Solution



I tried using the substitutions [itex]u=\sqrt[3]{1-x^7}[/itex] and [itex]u=\sqrt[7]{1-x^3}[/itex] 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
 

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