Trig Substitution Homework: Int x^2 + 36/4x^2

[mathor345]

Homework Statement

$$\int\frac{\sqrt{x^2 + 36}}{4x^2}}dx$$

Homework Equations

sqrt(a^2 + x^2) substitution for x = a tan theta

The Attempt at a Solution

I set
x = 6 tan theta
x^2 = 36 tan^2 theta
dx = 6 sec^2 x

$$\int\frac{\sqrt{36 + 36 tan \theta}}{144 tan \theta}dx$$

$$\int\frac{\sqrt{36(tan \theta + 1)}}{144 tan \theta}dx$$

$$\int\frac{\sqrt{36(sec^2 \theta)}}{144 tan \theta}dx$$

Then the numerator becomes 6 sec theta and then i multiply it by dx to get 36 sec^3 theta over 144 tan theta. Not sure where I went wrong. I've done other similar problems but they had the radical on bottom or the denominator was simply a variable.

 

[Char. Limit]

Firstly, the bottom should be 144 tan^2(theta). Squared.

Next, there isn't a problem here. Some trig sub problems are harder than others. This is one of them. I recommend converting the secant and tangent functions to their sine and cosine equivalents. It might make it easier.

EDIT: After you do that, the trig identity $$csc^2(\theta) = cot^2(\theta)+1$$ will be a big help.

 

[dextercioby]

$x= 6\sinh t$ as a substitution immediately solves the problem.