If ∫ 1/(a² sin² x + b² cos² x) dx = 1/12 tan⁻¹(3 tan x) + constant, then the maximum value of a sin

Q: Wait, why do we divide by $\cos^2 x$ at the start? Is there a rule for this?

That is a fantastic question! In JEE Advanced, whenever you encounter an integral where the denominator is a homogeneous quadratic expression in $\sin x$ and $\cos x$ (like $a^2 \sin^2 x + b^2 \cos^2 x$), dividing by $\cos^2 x$ is the standard 'golden key'. It immediately converts the denominator into a quadratic in $\tan x$ and the numerator into $\sec^2 x$, which is the derivative of $\tan x$. This is a powerful technique that turns a complex trigonometric integral into a simple algebraic one.

Q: How do we know $a$ and $b$ are positive? Could they be negative?

In the context of this problem, we are looking for the maximum value of $a \sin x + b \cos x$. Since the maximum value is $\sqrt{a^2 + b^2}$, the sign of $a$ and $b$ does not actually affect the final result because they are squared. However, by comparing our derived integral $\frac{1}{ab} \tan^{-1}(\frac{a \tan x}{b})$ with the given $\frac{1}{12} \tan^{-1}(3 \tan x)$, we can deduce that $a$ and $b$ must be positive to match the coefficients directly. It is a safe and logical assumption here.

Q: Is there a shortcut for the maximum value of $a \sin x + b \cos x$?

Yes, there is! You can think of this as a vector addition. The expression $a \sin x + b \cos x$ can be written as $\sqrt{a^2 + b^2} (\frac{a}{\sqrt{a^2 + b^2}} \sin x + \frac{b}{\sqrt{a^2 + b^2}} \cos x)$. If you let $\cos \phi = \frac{a}{\sqrt{a^2 + b^2}}$ and $\sin \phi = \frac{b}{\sqrt{a^2 + b^2}}$, this becomes $\sqrt{a^2 + b^2} \sin(x + \phi)$. Since the maximum value of the sine function is $1$, the maximum value of the entire expression is simply $\sqrt{a^2 + b^2}$. It is a beautiful application of trigonometry and geometry!

Question

If $\int \frac{1}{a^2 \sin^2 x + b^2 \cos^2 x} dx = \frac{1}{12} 	an^{-1}(3 	an x) + 	ext{constant}$, then the maximum value of $a \sin x + b \cos x$, is :

Options:

(A) $\sqrt{40}$

(B) $\sqrt{41}$

(C) $\sqrt{39}$

(D) $\sqrt{42}$

Sigma Percentile · Accepted Answer

### The Art of Transformation: Unlocking the Integral

Welcome, fellow traveler on the JEE Advanced journey. Today, we are going to dissect a problem that might look like a daunting wall of trigonometry, but it is actually a beautifully orchestrated dance of calculus.

We are given the integral:

$$ \int \frac{1}{a^2 \sin^2 x + b^2 \cos^2 x} dx = \frac{1}{12} 	an^{-1}(3 	an x) + C $$

Our mission is to find the constants $a$ and $b$ and then find the maximum value of $a \sin x + b \cos x$. Let us break this down, step by step.

### Phase 1: The Geometric Key

When you see an integral with $\sin^2 x$ and $\cos^2 x$ in the denominator, your first instinct should be to simplify. We want to move from the world of sines and cosines into the world of tangents.

We divide both the numerator and the denominator by $\cos^2 x$. Since $\frac{1}{\cos^2 x} = \sec^2 x$ and $\frac{\sin^2 x}{\cos^2 x} = 	an^2 x$, this transformation is the 'geometric key' that unlocks the entire problem.

The integral becomes:

$$ \int \frac{\sec^2 x}{a^2 	an^2 x + b^2} dx $$

### Phase 2: The Bridge to Algebra

Now that we have $\sec^2 x$ in the numerator, we have a gift. We know that the derivative of $	an x$ is $\sec^2 x$, which is the perfect setup for a substitution.

Let $t = 	an x$. Then, $dt = \sec^2 x dx$. Our integral transforms into a much friendlier algebraic form:

$$ \int \frac{dt}{a^2 t^2 + b^2} $$

To solve this, we factor out $a^2$ from the denominator:

$$ \frac{1}{a^2} \int \frac{dt}{t^2 + (b/a)^2} $$

Using the standard formula $\int \frac{dx}{x^2 + k^2} = \frac{1}{k} 	an^{-1}(\frac{x}{k}) + C$, where $k = \frac{b}{a}$, we obtain:

$$ \frac{1}{a^2} \cdot \frac{1}{b/a} 	an^{-1}\left(\frac{t}{b/a}ight) + C = \frac{1}{ab} 	an^{-1}\left(\frac{at}{b}ight) + C $$

Substituting $t = 	an x$ back, we get:

$$ \frac{1}{ab} 	an^{-1}\left(\frac{a 	an x}{b}ight) + C $$

### Phase 3: The Comparison

We have our result: $\frac{1}{ab} 	an^{-1}(\frac{a 	an x}{b})$. The problem gives us $\frac{1}{12} 	an^{-1}(3 	an x)$.

By comparing these two, we equate the coefficients:

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

Substituting $a = 3b$ into $ab = 12$, we get $3b^2 = 12$, which implies $b^2 = 4$. Assuming positive constants, we find $b = 2$ and $a = 6$.

### Phase 4: The Final Peak

We have found our constants: $a = 6$ and $b = 2$. The problem asks for the maximum value of $a \sin x + b \cos x$, which is $6 \sin x + 2 \cos x$.

The maximum value of any expression $p \sin x + q \cos x$ is given by $\sqrt{p^2 + q^2}$. Plugging in our values:

$$ \sqrt{6^2 + 2^2} = \sqrt{36 + 4} = \sqrt{40} $$

The final answer is **$\sqrt{40}$** (or $2\sqrt{10}$). You have navigated the transformation, the substitution, the comparison, and the final maximization. This is the essence of JEE Advanced mathematics—taking a complex problem and breaking it down into logical steps.

Animated Solution for Mathematics - Indefinite Integration: If $\int \frac{1}{a ^{2} s i n ^{2} x + b ^{2} c o s ^{2} x} d x = \frac{1}{12} tan^{- 1} (3 tan x) + constant$ , then the maximum value of $a sin x + b cos x$ , is :

Select Answer:

Visualized Solution

Analyze the Given Integral

The Strategy: Divide by $cos^{2} x$

Rewriting the Integral

Applying Substitution

Standard Integration Formula

Evaluating the Integral

Substituting Back

Comparing with Given Result

Solving for a and b

The Maximum Value Concept

Visualizing the Maximum

Final Calculation

The Sigma Insight: Evaluation of Special Integral Forms

The Art of Transformation

Phase 1

Phase 2

Phase 3

Phase 4

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