My calculus textbook threw me a curveball in evaluating an integral that I don't understand.
They reduced a constant plus a sum within a radical to just the sum within the radical.
1 + ((x^3)/2 - 1/2x^3)^2 = ((x^3)/2 + 1/2x^3)^2
This looks like voodoo magic.... where did the 1 go and what is the proper derivation of this procedure? I will need to repeat this operation so I need to know how it works.
Thanks.
They reduced a constant plus a sum within a radical to just the sum within the radical.
1 + ((x^3)/2 - 1/2x^3)^2 = ((x^3)/2 + 1/2x^3)^2
This looks like voodoo magic.... where did the 1 go and what is the proper derivation of this procedure? I will need to repeat this operation so I need to know how it works.
Thanks.
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If you expand left hand side you can see the result coming:
LHS = 1+〖(x^3/2-1/(2x^3 ))〗^2
……= 1+〖(x^3/2)〗^2 - 2.x^3/2.1/(2x^3 ) +〖(1/(2x^3 ))〗^2
……= 1 + 〖(x^3/2)〗^2 - 1/2 + 〖(1/(2x^3 ))〗^2
……=〖(x^3/2)〗^2 + 1/2 + 〖(1/(2x^3 ))〗^2 ….. see how the expression now has 1/2 not - 1/2
…..=〖 (x^3/2+ 1/(2x^3 ))〗^2
…..=RHS
LHS = 1+〖(x^3/2-1/(2x^3 ))〗^2
……= 1+〖(x^3/2)〗^2 - 2.x^3/2.1/(2x^3 ) +〖(1/(2x^3 ))〗^2
……= 1 + 〖(x^3/2)〗^2 - 1/2 + 〖(1/(2x^3 ))〗^2
……=〖(x^3/2)〗^2 + 1/2 + 〖(1/(2x^3 ))〗^2 ….. see how the expression now has 1/2 not - 1/2
…..=〖 (x^3/2+ 1/(2x^3 ))〗^2
…..=RHS