Key Concepts and Formulas

• Trigonometric Identities: $\sec^2 x = 1 + \tan^2 x$, $\sin^2 x + \cos^2 x = 1$
• Integration Formula: $\int \frac{1}{a^2 + x^2} dx = \frac{1}{a} \tan^{-1}\left(\frac{x}{a}\right) + C$
• Maximum Value: The maximum value of $a \sin x + b \cos x$ is $\sqrt{a^2 + b^2}$.

Step-by-Step Solution

Step 1: Manipulate the integral

We start with the given integral and manipulate it to a form that's easier to integrate. We divide both the numerator and the denominator by $\cos^2 x$. This allows us to express the integral in terms of $\tan x$.

$I = \int \frac{1}{a^2 \sin^2 x + b^2 \cos^2 x} dx = \int \frac{\frac{1}{\cos^2 x}}{\frac{a^2 \sin^2 x}{\cos^2 x} + \frac{b^2 \cos^2 x}{\cos^2 x}} dx = \int \frac{\sec^2 x}{a^2 \tan^2 x + b^2} dx$

Step 2: Perform a substitution

We substitute $t = \tan x$, which implies $dt = \sec^2 x dx$. This simplifies the integral further.

$t = \tan x \Rightarrow dt = \sec^2 x dx$ $I = \int \frac{dt}{a^2 t^2 + b^2}$

Step 3: Evaluate the integral

We use the standard integral formula $\int \frac{1}{a^2 + x^2} dx = \frac{1}{a} \tan^{-1}\left(\frac{x}{a}\right) + C$.

$I = \int \frac{dt}{a^2 t^2 + b^2} = \frac{1}{a^2} \int \frac{dt}{t^2 + \left(\frac{b}{a}\right)^2} = \frac{1}{a^2} \cdot \frac{1}{\frac{b}{a}} \tan^{-1}\left(\frac{t}{\frac{b}{a}}\right) + C = \frac{1}{ab} \tan^{-1}\left(\frac{at}{b}\right) + C$

Step 4: Substitute back for $t$

We substitute back $t = \tan x$ to obtain the integral in terms of $x$.

$I = \frac{1}{ab} \tan^{-1}\left(\frac{a \tan x}{b}\right) + C$

Step 5: Compare with the given result

We are given that $I = \frac{1}{12} \tan^{-1}(3 \tan x) + C$. Comparing this with our result, we have:

$\frac{1}{ab} = \frac{1}{12} \Rightarrow ab = 12$ $\frac{a}{b} = 3 \Rightarrow a = 3b$

Step 6: Solve for $a$ and $b$

Substituting $a = 3b$ into $ab = 12$, we get:

$(3b)b = 12 \Rightarrow 3b^2 = 12 \Rightarrow b^2 = 4 \Rightarrow b = 2$ Since $a = 3b$, we have $a = 3(2) = 6$.

Step 7: Calculate the maximum value of $a \sin x + b \cos x$

The maximum value of $a \sin x + b \cos x$ is given by $\sqrt{a^2 + b^2}$.

$\text{Maximum value} = \sqrt{a^2 + b^2} = \sqrt{6^2 + 2^2} = \sqrt{36 + 4} = \sqrt{40}$

Common Mistakes & Tips

• Careless Substitution: Be careful when substituting back for $t$. Ensure you replace it with the correct expression in terms of $x$.
• Incorrect use of Integration Formula: Remember the constant factor $\frac{1}{a}$ when using the integration formula for $\int \frac{1}{a^2 + x^2} dx$.
• Not simplifying the integral: Always simplify the integral as much as possible before attempting to integrate.

Summary

We started by manipulating the given integral using trigonometric identities and a suitable substitution. We then evaluated the integral and compared it with the given result to find the values of $a$ and $b$. Finally, we used these values to calculate the maximum value of $a \sin x + b \cos x$, which is $\sqrt{40}$.

The final answer is \boxed{\sqrt{40}}, which corresponds to option (C).