Ignoring positive/negative values with trig substitutions?

[process91]

If I wanted to integrate $\int \sqrt{1+x^2} dx$, I would let $x=\tan\theta$ , which implies $dx=\sec^2 \theta dx$ so that I would have:

$\int \sqrt{1+x^2} dx = \int \sqrt{1 + \tan^2 \theta} \sec \theta d \theta = \int \sqrt{\sec^2 \theta}\sec^2\theta d\theta = \int \sec^3 \theta d \theta$

It is this last equality that I am questioning. Why is it not $\int |\sec \theta| \sec^2 \theta d \theta$?

 

[LCKurtz]

If I wanted to integrate $\int \sqrt{1+x^2} dx$, I would let $x=\tan\theta$ , which implies $dx=\sec^2 \theta dx$ so that I would have:

$\int \sqrt{1+x^2} dx = \int \sqrt{1 + \tan^2 \theta} \sec \theta d \theta = \int \sqrt{\sec^2 \theta}\sec^2\theta d\theta = \int \sec^3 \theta d \theta$

It is this last equality that I am questioning. Why is it not $\int |\sec \theta| \sec^2 \theta d \theta$?

Any value of x can be gotten by that substitution with -π/2 < θ < π/2. sec(θ) is positive in that domain.

 

[process91]

Thanks, that does clear it up a lot. So for all trig substitutions, then, there is an implied statement about the range of values which x can take and the corresponding range of values provided by the substitution?