(A) (-1, 1 - tanθ) (B) (-1, 1 + tanθ) (C) (1, 1 + tanθ) (D) (1, 1 - tanθ)

Q: Why can't we just use $\sin\theta$ and $\cos\theta$ instead of converting to $\tan\theta$?

You *could*, but it would be like trying to solve a maze by walking through walls. Converting to $\tan\theta$ is the 'path of least resistance' because $\sec^2\theta d\theta$ is the derivative of $\tan\theta$. It turns a complex trigonometric mess into a simple polynomial division.

Q: Why did we simplify the denominator *before* substituting?

Great question! If you substitute $t = \tan\theta$ *before* simplifying the denominator, you end up with a nested fraction that is much harder to manage. Simplifying first is like cleaning your workspace before you start building—it makes the actual integration step trivial.

Q: Is the constant $C$ really necessary? Can't we just ignore it?

In JEE Advanced, precision is everything. The constant $C$ represents the 'family' of solutions. Ignoring it is like ignoring air resistance in a physics problem—it might seem small, but it changes the fundamental nature of the answer. Always include it!

Question

If $\int \frac{d	heta}{\cos^2	heta (	an 2	heta + \sec 2	heta)} = \lambda 	an	heta + 2 \log_e |f(	heta)| + C$ where C is constant if integration, then the ordered pair $(\lambda, f(	heta))$ is equal to:

Options:

(A) $(-1, 1 - 	an	heta)$

(B) $(-1, 1 + 	an	heta)$

(C) $(1, 1 + 	an	heta)$

(D) $(1, 1 - 	an	heta)$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing before a complex integral:

$$I = \int \frac{d	heta}{\cos^2	heta (	an 2	heta + \sec 2	heta)}$$

It looks intimidating, but in the world of JEE Advanced, every problem is a puzzle waiting to be solved. The first step is to look for the 'hidden key.'

Notice that $\frac{1}{\cos^2	heta}$ is a massive neon sign pointing towards a specific substitution. Since $\frac{1}{\cos^2	heta} = \sec^2	heta$ and $\sec^2	heta$ is the derivative of $	an	heta$, we have our roadmap. We rewrite the integral as:

$$I = \int \frac{\sec^2	heta \, d	heta}{	an 2	heta + \sec 2	heta}$$

### The Trigonometric Bridge: Double Angle Identities

Now, look at the denominator. We have $	an 2	heta$ and $\sec 2	heta$, which are double-angle terms currently blocking our path. To clear the way, we bring them into the world of $	an	heta$ using the classic identities:

$$	an 2	heta = \frac{2	an	heta}{1-	an^2	heta} \quad 	ext{and} \quad \sec 2	heta = \frac{1+	an^2	heta}{1-	an^2	heta}$$

By substituting these into our denominator $D$, we get:

$$D = \frac{2	an	heta}{1-	an^2	heta} + \frac{1+	an^2	heta}{1-	an^2	heta} = \frac{1 + 	an^2	heta + 2	an	heta}{1-	an^2	heta}$$

### The Algebraic Transformation

Notice that the numerator of our new denominator, $1 + 	an^2	heta + 2	an	heta$, is the perfect square expansion of $(1+	an	heta)^2$. Thus, the denominator becomes:

$$D = \frac{(1+	an	heta)^2}{1-	an^2	heta}$$

Substituting this back into our integral, we obtain:

$$I = \int \frac{\sec^2	heta (1-	an^2	heta)}{(1+	an	heta)^2} \, d	heta$$

Now, let $t = 	an	heta$, which implies $dt = \sec^2	heta \, d	heta$. Our integral transforms into a simple algebraic rational function:

$$I = \int \frac{1-t^2}{(1+t)^2} \, dt$$

### The Final Integration

Factor the numerator as $1-t^2 = (1-t)(1+t)$. Our integral becomes:

$$I = \int \frac{(1-t)(1+t)}{(1+t)^2} \, dt = \int \frac{1-t}{1+t} \, dt$$

To integrate this, we perform a clever manipulation:

$$\frac{1-t}{1+t} = \frac{-(t+1) + 2}{1+t} = -1 + \frac{2}{1+t}$$

Integrating term by term, we find:

$$I = -t + 2 \ln |1+t| + C$$

Finally, substituting $t = 	an	heta$ back into the expression, we get:

$$I = -	an	heta + 2 \ln |1+	an	heta| + C$$

Comparing this to the form $\lambda 	an	heta + 2 \ln |f(	heta)| + C$, we identify **$\lambda = -1$** and **$f(	heta) = 1 + 	an	heta$**. The final result is **$(\lambda, f(	heta)) = (-1, 1 + 	an	heta)$**.

Animated Solution for Mathematics - Indefinite Integration: If $\int \frac{d θ}{c o s ^{2} θ ( t a n 2 θ + s e c 2 θ )} = λ tan θ + 2 lo g_{e} ∣ f (θ) ∣ + C$ where C is constant if integration, then the ordered pair $(λ, f (θ))$ is equal to:

Select Answer:

Visualized Solution

Analyze the Integrand Structure

Recall Double Angle Identities

Substitute Identities into Denominator

Simplify the Denominator Expression

Apply Substitution $t = tan θ$

Algebraic Factorization

Manipulate the Fraction for Integration

Perform the Integration

Back-Substitution and Comparison

Final Conclusion

The Sigma Insight: Integration by Substitution

Analyzing the Setup

The Trigonometric Bridge

The Algebraic Transformation

The Final Integration

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