Limits on theta in trigonometric substitution

[susume]

Hello,
I am trying to understand how the domain of theta is determined for the trigonometric substitutions when integrating. In the case of an integrand containing sqrt(x^2-a^2), when we set x=a*sec(theta), what is the domain of theta and why? My textbook's online tutorial says theta is then limited to the first and third quadrants, which sort of makes sense to me because sqrt(x^2-a^2)=a*tan(theta) is positive and these are the quadrants where tan(theta) is positive. My instructor however says in this case theta is limited to the first and second quadrants, and sqrt(x^2-a^2)= plus or minus a*tan(theta) rather than simply a*tan(theta). Can someone shed light on this for me?

I see that in the other trig substitutions, theta is limited to the first and 4th quadrants. Is this because that is where sqrt(a^2-x^2)=a*cos(theta) and sqrt(a^2+x^2)=a*sec(theta) are positive, or for some other reason? In the case of a definite integral, does one have to somehow check the upper and lower bounds of x against the permitted domain of theta?

Thank you for your time and help!

 

[Deleted member 4993]

Hello,
I am trying to understand how the domain of theta is determined for the trigonometric substitutions when integrating. In the case of an integrand containing sqrt(x^2-a^2), when we set x=a*sec(theta), what is the domain of theta and why? My textbook's online tutorial says theta is then limited to the first and third quadrants, which sort of makes sense to me because sqrt(x^2-a^2)=a*tan(theta) is positive and these are the quadrants where tan(theta) is positive. My instructor however says in this case theta is limited to the first and second quadrants, and sqrt(x^2-a^2)= plus or minus a*tan(theta) rather than simply a*tan(theta). Can someone shed light on this for me?

I see that in the other trig substitutions, theta is limited to the first and 4th quadrants. Is this because that is where sqrt(a^2-x^2)=a*cos(theta) and sqrt(a^2+x^2)=a*sec(theta) are positive, or for some other reason? In the case of a definite integral, does one have to somehow check the upper and lower bounds of x against the permitted domain of theta?

Thank you for your time and help!

\(\displaystyle \theta \ = \ sec^{-1}(x)\)

The range of Θ is π ≥ Θ ≥ 0.

To look at domain and ranges of different inverse-trig functions - do a google search.