The question asks which of the following integrals is easily integrated by substituting $x = 6 \tan \theta$. The given options are: A. $\int \frac{1}{\sqrt{36+x^2}} dx$ B. $\int \frac{1}{\sqrt{36-x^2}} dx$ C. $\int \frac{1}{\sqrt{6-x^2}} dx$ D. $\int \frac{1}{\sqrt{6+x^2}} dx$

1. Problem Description

The question asks which of the following integrals is easily integrated by substituting $x = 6 \tan \theta$. The given options are:

A. $\int \frac{1}{\sqrt{36+x^2}} dx$

B. $\int \frac{1}{\sqrt{36-x^2}} dx$

C. $\int \frac{1}{\sqrt{6-x^2}} dx$

D. $\int \frac{1}{\sqrt{6+x^2}} dx$

2. Solution Steps

We are given the substitution $x = 6 \tan \theta$. This suggests that $x^2 = 36 \tan^2 \theta$. Let's examine how this substitution would simplify each integral:

A. $\int \frac{1}{\sqrt{36+x^2}} dx = \int \frac{1}{\sqrt{36 + 36 \tan^2 \theta}} dx = \int \frac{1}{\sqrt{36(1 + \tan^2 \theta)}} dx = \int \frac{1}{6\sqrt{\sec^2 \theta}} dx = \int \frac{1}{6 \sec \theta} dx$. Since $x = 6 \tan \theta$, $dx = 6 \sec^2 \theta d\theta$. Therefore, the integral becomes $\int \frac{1}{6 \sec \theta} 6 \sec^2 \theta d\theta = \int \sec \theta d\theta$, which is a standard integral.

B. $\int \frac{1}{\sqrt{36-x^2}} dx = \int \frac{1}{\sqrt{36 - 36 \tan^2 \theta}} dx = \int \frac{1}{\sqrt{36(1 - \tan^2 \theta)}} dx$. The expression $1 - \tan^2 \theta$ does not simplify nicely.

C. $\int \frac{1}{\sqrt{6-x^2}} dx$. Here, the constant term is 6, not 36, so $x = 6 \tan \theta$ is not the appropriate substitution.

D. $\int \frac{1}{\sqrt{6+x^2}} dx$. Here, the constant term is 6, not 36, so $x = 6 \tan \theta$ is not the appropriate substitution.

We can analyze the first option in more detail:

If $x = 6 \tan \theta$, then $dx = 6 \sec^2 \theta d\theta$. Then $\sqrt{36 + x^2} = \sqrt{36 + 36 \tan^2 \theta} = \sqrt{36(1 + \tan^2 \theta)} = \sqrt{36 \sec^2 \theta} = 6 \sec \theta$. Thus, the integral becomes

$\int \frac{1}{\sqrt{36+x^2}} dx = \int \frac{1}{6 \sec \theta} 6 \sec^2 \theta d\theta = \int \sec \theta d\theta = \ln|\sec \theta + \tan \theta| + C$. Since $x = 6 \tan \theta$, we have $\tan \theta = \frac{x}{6}$. Then $\sec \theta = \sqrt{1 + \tan^2 \theta} = \sqrt{1 + \frac{x^2}{36}} = \sqrt{\frac{36 + x^2}{36}} = \frac{\sqrt{36 + x^2}}{6}$. Therefore, the integral is $\ln|\frac{\sqrt{36 + x^2}}{6} + \frac{x}{6}| + C = \ln|\sqrt{36 + x^2} + x| - \ln 6 + C = \ln|\sqrt{36 + x^2} + x| + C'$, which is a valid result.

3. Final Answer

A. $\int \frac{1}{\sqrt{36+x^2}} dx$