Question

Integrals with hyperbolic/trigonometric substitutions????

Hi, so I've got an MEI FP2 mock tomorrow, but I'm still not sure how to do this.

If you've got an integral like:

`int_3^5sqrt(x^2-9)dx`

or

`int_0^1 (2x)/sqrt(x^2-9)dx`

The second example is meant to be a trigonometric/hyperbolic integration (fairly standard), but with a 2x instead of just a 1. I have just made that second integral up, I can't seem to find any at the moment, but they seem to integrate to something like `asqrt(x^2-9)` where a is some number that to me has just been magicked out of thin air???

If anyone can help, that'd be great!!

P.S. The first integral, I just get stuck on what to substitute

Answer 1

Some ideas...

Explanation:

Here are a couple of ideas to (hopefully) get you unstuck...

For:

`int_3^5 sqrt(x^2-9) dx`

Note that `x` ranges from `3` to `5`.

We could try putting `x = 3sec u`. Then `dx/(du) = 3sec u tan u`

Then:

`int sqrt(x^2-9) dx = int sqrt(9sec^2 u-9) dx/(du) du`

`color(white)(int sqrt(x^2^9) dx) = int sqrt(9tan^2 u) (3 sec u tan u) du`

`color(white)(int sqrt(x^2^9) dx) = int (9 sec u tan^2 u) du`

That's still quite yucky.

Let's try a hyperbolic substitution instead:

Put `x = 3 cosh u`.

Then:

`x+sqrt(x^2-9) = 3e^u`

So:

`u = ln(1/3(x+sqrt(x^2-9)))`

Also:

`dx/(du) = 3 sinh u`

and:

`int sqrt(x^2-9)color(white)(.)dx = int sqrt(9 cosh^2 u - 9) dx/(du) du`

`color(white)(int sqrt(x^2-9).dx) = int sqrt(9 sinh^2 u) (3 sinh u) du`

`color(white)(int sqrt(x^2-9).dx) = int (9 sinh^2 u) du`

`color(white)(int sqrt(x^2-9).dx) = int 9/2 (cosh 2u - 1) du`

`color(white)(int sqrt(x^2-9).dx) = 9/4 sinh 2u - 9/2 u + color(grey)("constant")`

`color(white)(int sqrt(x^2-9).dx) = 1/2x sqrt(x^2-9) - 9/2ln(1/3(x+sqrt(x^2-9))) + color(grey)("constant")`

`color(white)(int sqrt(x^2-9).dx) = 1/2x sqrt(x^2-9) - 9/2ln(x+sqrt(x^2-9)) + color(grey)("constant")`

For the second integral, note that `2x` is the derivative of `(x^2-9)`.

So we don't need to use a substitution at all, since:

`d/(dx) sqrt(x^2-9) = (2x)/sqrt(x^2-9)`

So:

`int (2x)/(sqrt(x^2-9)) dx = sqrt(x^2-9) + color(grey)("constant")`

Notes

The strange thing about the second integral is the requested range is `0` to `1`. So the square root in the denominator will be non-real. It is still correct - just a bit strange.

Should it have been `sqrt(9-x^2)` or `sqrt(x^2+9)` ?

The reason I mention this is that the range of a definite integral may inform whether to use a trigonometric `abs|...| <= 1` or hyperbolic `abs|...| >= 1` substitution.

For something like `int (2x^2)/sqrt(x^2-9) dx` you could try `x = 3 sec u` or `x = 3 cosh u`, but for int `(2x^2)/sqrt(9-x^2) dx` you might try `x = 3 cos u` if the range of integration was appropriate.