Evaluating Integrals Using Trigonometric Function Substitutions Question

[GreenPrint]

Hi,

I just had this idea pop into my head... Can you use a trig sub with a reference triangle who has sides equal to zero? or more like a value close to zero such as dx or da or something?

For example

integral 1/sqrt(9+dx^2) (dx)^2

would have a reference triangle were the hypotenuse is sqrt(9+dx^2)
one of the legs would be 3
the other leg would be dx

you then could express dx/sqrt(9+dx^2) as either csc(theta) or sec(theta) depending on what leg you set 3 be equal to...

integral 1/sqrt(9+dx^2) (dx)^2 = integral csc(theta) dx

 

[GreenPrint]

I really wounder what can be done if one sets the side of the length of the triangle to dx and try and make a sub some how if that makes any sense

 

[HallsofIvy]

Hi,

I just had this idea pop into my head... Can you use a trig sub with a reference triangle who has sides equal to zero? or more like a value close to zero such as dx or da or something?

For example

integral 1/sqrt(9+dx^2) (dx)^2

This is an invalid integral. You cannot treat the "dx" as if it were a variable.
An integral in x must be of the form $\int f(x)dx$.

would have a reference triangle were the hypotenuse is sqrt(9+dx^2)
one of the legs would be 3
the other leg would be dx

you then could express dx/sqrt(9+dx^2) as either csc(theta) or sec(theta) depending on what leg you set 3 be equal to...

integral 1/sqrt(9+dx^2) (dx)^2 = integral csc(theta) dx