# Trigonometric substitution problem

• April 20th 2010, 10:01 AM
s3a
Trigonometric substitution problem
The initial problem is: http://www.wolframalpha.com/input/?i=integrate(sqrt(x^2+-+1)%2Fx%2C+x%2C1%2C2)

My problem is not in the substitution part but rather on the limits of integration. I tried "an easy way out" which is to just use => instead of = signs and not put limits of integration and then just turn the variables into x variables (initial variables) and use the initial limits of integration but the problem I am having is from this step:

integral of (tan^2 (theta) (dtheta)) = tan(theta) - (theta) + C = sqrt(x^2 - 1) - arcsec(x) and when I try to find the value for the final answer:

sqrt(2^2 - 1) - arcsec(2) - 0 + arcsec(1)

both bold parts give me errors on the calculator which means they do not exist. I do recall that theta should be between [0, pi/2) or [pi, 3pi/2) but there is a final answer in the back of the book (sqrt(3) - pi/3)which suggests I am doing something wrong.

Any help would be GREATLY appreciated!
• April 20th 2010, 11:02 AM
HallsofIvy
Quote:

Originally Posted by s3a
The initial problem is: http://www.wolframalpha.com/input/?i=integrate(sqrt(x^2+-+1)%2Fx%2C+x%2C1%2C2)

My problem is not in the substitution part but rather on the limits of integration. I tried "an easy way out" which is to just use => instead of = signs and not put limits of integration and then just turn the variables into x variables (initial variables) and use the initial limits of integration but the problem I am having is from this step:

integral of (tan^2 (theta) (dtheta)) = tan(theta) - (theta) + C = sqrt(x^2 - 1) - arcsec(x) and when I try to find the value for the final answer:

sqrt(2^2 - 1) - arcsec(2) - 0 + arcsec(1)

both bold parts give me errors on the calculator which means they do not exist. I do recall that theta should be between [0, pi/2) or [pi, 3pi/2) but there is a final answer in the back of the book (sqrt(3) - pi/3)which suggests I am doing something wrong.

Any help would be GREATLY appreciated!

Are you sure you took "arcsec" rather than "arcsin" or "arccos"? The sine and cosine values are always between -1 and 1 so arcsin(2) or arccos(2) do not exist (though I can't imagine why "arcsin(1)" or "arccos(1)" would be a problem) but since secant is "1 over cosine" arcsec(2) is the same as arccos(1/2) which is $\pi/3$ and, of course, arcsec(1) is the same as arccos(1) which is 0.
• April 20th 2010, 12:50 PM
s3a
I had tried arcsec(2) = 1/arccos(2) and arcsec(1) = 1/arccos(1). So, what I have to do instead is take the reciprocal of the fraction and then use the reciprocal trig?

By this I mean, on the first example, arcsec(2) = arcsec(2/1) = arccos(1/2) ?

and for the second;

arcsec(1/1) = arccos(1/1) ?
• April 20th 2010, 03:36 PM
ione
Quote:

Originally Posted by s3a
I had tried arcsec(2) = 1/arccos(2) and arcsec(1) = 1/arccos(1). So, what I have to do instead is take the reciprocal of the fraction and then use the reciprocal trig?

By this I mean, on the first example, arcsec(2) = arcsec(2/1) = arccos(1/2) ?

and for the second;

arcsec(1/1) = arccos(1/1) ?

I find it helpful to think of arcsec(2) as the angle whose secant is 2. So if the secant of the angle is 2 then the cosine of the angle is 1/2 and

arcsec(2) = arccos(1/2)