Calculus BC Trigonometric Substitition

[hwmaltby]

Homework Statement

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

Homework Equations

$$\sin^2{\theta}+\cos^2{\theta}=1$$

The Attempt at a Solution

$$x=\tan{\theta}$$

I simplified it down to:

$$\int \frac {1}{\sin{\theta} \cdot \cos^2{\theta}} \, d\theta$$

Which I do not know how to integrate.

Any help would be wonderful!

 

[hwmaltby]

Never mind... I rewrote it as:

$$\int \csc{\theta}\cdot\sec^2{\theta} \, d\theta$$

And used u-substition:

$$u=\sec{\theta}$$

$$\int \frac{\csc{\theta} \cdot \sec^2{\theta}}{\tan{\theta} \cdot \sec{\theta}} \, du$$

$$\int \csc^2{\theta} \, du$$

$$\int \frac{u^2}{u^2-1} \, du$$

$$\frac{1}{2} \cdot \ln |\frac{u-1}{u+1}| +u+C$$

I then simplified using my previous substitutions and logarithm rules to get my final answer:

$$\ln |\frac{\sqrt{1+x^2}-1}{x}|+\sqrt{1+x^2}+C$$

 

[Mark44]

And you can check by differentiating your answer. If you end up with sqrt(1 + x^2)/x, you're golden.