I have a question about defining domain when using trig substitution.
When we integrate some function defined for all real values, do we have to define domain even though I am not going to use inverse trig functions? (However, in case of my question, $x!=0$ and it is a definite integral.)
In this case, $$\int_0^\infty \frac x{x^2+1} dx$$ I could've done $$\ u = \ln(x^2+1) $$ But I did the trig substitution$\ x = \tan \delta $ without defining its domain.
Using some algebra I got $$\lim\limits_{b \to \infty}\int_0^b \tan \delta \ d\delta \rightarrow \text{diverges}$$
My professor said that I have to define domain when I use trig sub, and that we should do the following. $$\lim\limits_{b \to \pi/2}\int_0^b \tan \delta \ d\delta \rightarrow \text{diverges}$$
However, I personally think if we don't use trig inverse function, I don't have to define domain when we use trig substitution.
Thank you
Any time you do any substitution -- trig or not -- you have to adjust the limits of integration. In your case, the limits for $x$ are $0$ to $\infty$, but the corresponding limits for $\delta$ will be from $\arctan 0 = 0$ to $\arctan \infty = \frac{\pi}2$.