Trigonometric Integral Problem

[Ahlahn]

Homework Statement

Evaluate the definite integral of function dx/x^2+1 on interval [tan(3), tan(6)]

Homework Equations

The Attempt at a Solution

The trigonometric inverse function for dx/x^2+1 is arctanx+c.
I plugged in tan(3) and tan(6), and subtracted arctan(tan(3)) from arctan(tan(6)) and got the answer -.1415. But it's incorrect...

What am I doing wrong?

 

[╔(σ_σ)╝]

arctan( tan(x)) = x

Use this.

Your calculator may be in radians or something.

 

[Ahlahn]

I switched it to degrees and it worked. Wow. Thanks! What's wrong with radians?!

 

[Dick]

arctan( tan(x)) = x

Use this.

Your calculator may be in radians or something.

Noo. The answer IS approximately -.1415. What's wrong is that arctan(tan(x))=x. That's only true if x is in (-pi/2,pi/2) which is the range of arctan. Otherwise arctan(tan(x)) will give you the number 'a' in (-pi/2,pi/2) such that tan(x)=tan(a). It will differ from x by a multiple of pi. -.1415 may be coming up wrong because it's not the exact answer. You can find an exact answer. Your 'interval' is also a little funny. tan(3)>tan(6). That's why your answer is negative.

 

[Dick]

I switched it to degrees and it worked. Wow. Thanks! What's wrong with radians?!

Ok. I was diagnosing the wrong problem. There is nothing wrong with radians. The only problem is 3 radians isn't equal to 3 degrees. The problem should have specified that 3 meant degrees.

 

[╔(σ_σ)╝]

Noo. The answer IS approximately -.1415. What's wrong is that arctan(tan(x))=x. That's only true if x is in (-pi/2,pi/2) which is the range of arctan. Otherwise arctan(tan(x)) will give you the number 'a' in (-pi/2,pi/2) such that tan(x)=tan(a). It will differ from x by a multiple of pi. -.1415 may be coming up wrong because it's not the exact answer. You can find an exact answer. Your 'interval' is also a little funny. tan(3)>tan(6). That's why your answer is negative.

Mistake of mine :-(. I gave a hurried answer without checking.