Let α ∈ (0, π/2) be fixed. If the integral ∫ (tan x + tan α) / (tan x - tan α) dx = A(x) cos 2α + B(

Q: Why can't we just integrate the numerator and denominator separately? Is there a quotient rule for integration?

That is a very common trap! Unlike differentiation, where we have the quotient rule, integration does not have a simple quotient rule. You cannot integrate the numerator and denominator separately and divide the results. You must simplify the integrand into a sum of terms that you can integrate individually. Always aim to transform the integrand into a form where standard integration formulas apply.

Q: I ended up with $x \cos 2\alpha + \sin 2\alpha \ln |\sin(x - \alpha)| + C$. But the options have $(x - \alpha) \cos 2\alpha$. Did I make a mistake?

You are actually on the right track! The difference between your result and the options is just a constant. Remember that $\alpha$ is a fixed constant, so $\alpha \cos 2\alpha$ is also a constant. When you expand $(x - \alpha) \cos 2\alpha$, you get $x \cos 2\alpha - \alpha \cos 2\alpha$. That extra term $-\alpha \cos 2\alpha$ simply gets absorbed into the arbitrary constant of integration $C$. In indefinite integration, constants are flexible!

Q: Why did we choose to rewrite $x + \alpha$ as $(x - \alpha) + 2\alpha$? Why not something else?

Great question. We chose $(x - \alpha) + 2\alpha$ because our denominator was $\sin(x - \alpha)$. By creating an $(x - \alpha)$ term in the numerator, we allow the $\sin(x - \alpha)$ term in the denominator to cancel out when we expand using the compound angle formula. It is a strategic move to simplify the expression, not a random choice. Always look at your denominator to decide how to manipulate your numerator.

Question

Let $\alpha \in (0, \pi/2)$ be fixed. If the integral $\int \frac{	an x + 	an \alpha}{	an x - 	an \alpha} dx = A(x) \cos 2\alpha + B(x) \sin 2\alpha + C$, where C is a constant of integration, then the functions A(x) and B(x) are respectively :

Options:

(A) $x - \alpha$ and $\log_e |\cos(x - \alpha)|$

(B) $x + \alpha$ and $\log_e |\sin(x - \alpha)|$

(C) $x - \alpha$ and $\log_e |\sin(x - \alpha)|$

(D) $x + \alpha$ and $\log_e |\sin(x + \alpha)|$

Sigma Percentile · Accepted Answer

### The Tangent Trap: A Journey into Trigonometric Integration

Welcome, student. Today, we are going to dismantle a problem that often intimidates students at first glance: the integral of a ratio of tangents. It looks messy, doesn't it?

$$I = \int \frac{	an x + 	an \alpha}{	an x - 	an \alpha} dx$$

But in the world of JEE Advanced, complexity is often just a mask for elegance. Let us peel back that mask together.

### Phase 1: The Sine-Cosine Bridge

Whenever you see an expression dominated by $	an x$, your first instinct should be to return to the roots: sine and cosine. Tangents are opaque; they hide the relationship between angles. Sines and cosines are transparent.

By substituting $	an 	heta = \frac{\sin 	heta}{\cos 	heta}$, we transform our integral into:

$$I = \int \frac{\frac{\sin x}{\cos x} + \frac{\sin \alpha}{\cos \alpha}}{\frac{\sin x}{\cos x} - \frac{\sin \alpha}{\cos \alpha}} dx$$

At this stage, it looks like a nightmare of fractions within fractions. But do not panic. This is the 'messy middle' that precedes every great simplification.

By taking the lowest common multiple (LCM) of $\cos x \cos \alpha$ in both the numerator and the denominator, the complex fraction collapses. The common denominator $\cos x \cos \alpha$ cancels out beautifully, leaving us with:

$$I = \int \frac{\sin x \cos \alpha + \cos x \sin \alpha}{\sin x \cos \alpha - \cos x \sin \alpha} dx$$

Look at that! The chaos has vanished, replaced by a structure that screams 'trigonometric identity'.

### Phase 2: The Art of Angle Manipulation

Now, we recognize the compound angle formulas. The numerator is the expansion of $\sin(x + \alpha)$, and the denominator is the expansion of $\sin(x - \alpha)$. So, our integral simplifies to:

$$I = \int \frac{\sin(x + \alpha)}{\sin(x - \alpha)} dx$$

Here is where the true test of a JEE aspirant lies. We have a mismatch: the numerator has $(x + \alpha)$, but the denominator has $(x - \alpha)$. We cannot integrate this directly.

We need to force the numerator to 'speak the language' of the denominator. We use the 'add and subtract' trick: rewrite $x + \alpha$ as $(x - \alpha) + 2\alpha$. Now, our integral becomes:

$$I = \int \frac{\sin((x - \alpha) + 2\alpha)}{\sin(x - \alpha)} dx$$

By treating $(x - \alpha)$ as a single block, we can expand the numerator using $\sin(A + B) = \sin A \cos B + \cos A \sin B$. This yields:

$$I = \int \frac{\sin(x - \alpha) \cos 2\alpha + \cos(x - \alpha) \sin 2\alpha}{\sin(x - \alpha)} dx$$

### Phase 3: The Final Integration and Constant Absorption

We are in the home stretch. Splitting the integral is now trivial. Dividing each term by $\sin(x - \alpha)$, we get:

$$I = \int (\cos 2\alpha + \cot(x - \alpha) \sin 2\alpha) dx$$

Since $\alpha$ is a constant, $\cos 2\alpha$ and $\sin 2\alpha$ are just constants. Integrating with respect to $x$, we obtain:

$$I = x \cos 2\alpha + \sin 2\alpha \ln |\sin(x - \alpha)| + C$$

But wait! The problem asks for the form $A(x) \cos 2\alpha + B(x) \sin 2\alpha + C$. Our first term is $x \cos 2\alpha$. To match the options, we rewrite $x$ as $(x - \alpha) + \alpha$.

Thus, $x \cos 2\alpha = (x - \alpha) \cos 2\alpha + \alpha \cos 2\alpha$. Since $\alpha \cos 2\alpha$ is a constant, we absorb it into the integration constant $C$.

The final form is:

$$(x - \alpha) \cos 2\alpha + \ln |\sin(x - \alpha)| \sin 2\alpha + C'$$

Comparing this to the target form, we identify **$A(x) = x - \alpha$** and **$B(x) = \ln |\sin(x - \alpha)|$**. We have arrived at the solution, not by brute force, but by understanding the geometric and algebraic soul of the problem. Keep this mindset, and no integral will ever be too daunting.

Animated Solution for Mathematics - Indefinite Integration: Let $α \in (0, π /2)$ be fixed. If the integral $\int \frac{t a n x + t a n α}{t a n x - t a n α} d x = A (x) cos 2 α + B (x) sin 2 α + C$ , where C is a constant of integration, then the functions A(x) and B(x) are respectively :

Select Answer:

Visualized Solution

The Given Integral

Converting to Sine and Cosine

Simplifying the Fraction

Applying Trigonometric Identities

Angle Manipulation Strategy

Expanding the Numerator

Splitting the Integral

Performing Integration

Adjusting the Constant

Final Comparison

Conclusion

The Sigma Insight: Integration by Substitution

The Tangent Trap

Phase 1

Phase 2

Phase 3

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