Question

Do the identities used in integration by trigonometric substitution matter?

Let's say I have a function `f(x) = sqrt(9-x^2)/x^2`

In order to evaluate `int f(x) dx` using trigonometric substitution, I can use the following identity ;

`sin^2(theta) + cos^2(theta) = 1`

and set `x = 3cos(theta)`, `dx = -3sin(theta) d theta`, with `sqrt(9-x^2) = 3sin(theta)`

Substituting, I get :

`int f(x)dx = -9/9int( sin^2(theta)/cos^2(theta))d theta = - int tan^2(theta)d theta`

`- int tan^2(theta)d theta = -int(sec^2(theta) -1)d theta = int d theta - int sec^2(theta) d theta = theta - tan(theta) + C`

When I express `theta` in terms of `x`, I can use `x/3 = cos(theta)` from my initial substitution, and then obtain `tan(theta) = sqrt(9 - x^2)/x`, which in turn gives me :

`int f(x)dx = cos^-1(x/3) - sqrt(9 - x^2)/x + C`

If I had chosen `x = 3sin(theta)` at the beginning, the integral becomes :

`int f(x)dx = - sqrt(9 - x^2)/x - sin^-1(x/3) + C`

So my question essentially boils down to: Does the choice of `x` at the beginning of the process matter? My intuition tells me that it shouldn't matter, yet I get different results when I compute both integrals (I've checked for mistakes and can't seem to find anything??)) I can also show my workings for the second integral if needed.

Answer 1

See below

Explanation:

Integration (as with differentiation) is unique (with the exception of constant). So if you get two solutions there are two possible inferences: (1) you made a mistake or (2) the two solutions are equivalent.

Because `sin^(−1)x+cos^(−1)(x)=π/2` the two solutions you have are the same but the constant `C` is different. Similarly, because `sinx` and `cosx` are identical curves, just phase shifted, then anytime you would substitute `sinx` then equally you can use `cosx` and get the same result.