Trig Sub. Integral: Limits of 1 to \sqrt{3}

[3.141592654]

Homework Statement

I have to find the definite integral with limits of integration of 1 to $$\sqrt{3}$$ for:

$$\int\frac{\sqrt{1+x^2}}{x^2}$$

Homework Equations

The Attempt at a Solution

I used trig. sub., so I have:

$$x=tan \theta$$

$$dx=(sec \theta)^2$$

So:

$$=\int\frac{\sqrt{1+(tan \theta)^2}}{(tan\theta)^2}(sec \theta)^2 d\theta$$

$$=\int\frac{\sqrt{(sec \theta)^2}}{(tan\theta)^2}(sec \theta)^2 d\theta$$

$$=\int\frac{(sec \theta)^3d\theta}{(tan\theta)^2}$$

I can play around with U-sub or Trig. identities but I'm missing something.

 

[n!kofeyn]

Well you can simplify the integrand to
$$csc^2 \theta \sec\theta = (1+\cot^2\theta)\sec\theta$$
From here, you should be able to break it down into known anti-derivatives.

 

[vela]

You may want to look as using the hyperbolic trig functions instead of the regular trig functions.

Or continue from where you are and try writing the integrand in terms of sine and cosine.

 

[n!kofeyn]

You may want to look as using the hyperbolic trig functions instead of the regular trig functions.

Well the above hint I gave gives very simple integrands to find the anti-derivatives of, so I don't think there's any reason to switch to hyperbolic trigonometric functions.

 

[vela]

Well the above hint I gave gives very simple integrands to find the anti-derivatives of, so I don't think there's any reason to switch to hyperbolic trigonometric functions.

There's more than one way to solve the problem, and it doesn't hurt to see how the various methods work out.