If b_n = ∫[0 to π/2] (cos² nx / sin x) dx, n ∈ N, then (A) b3 - b2, b4 - b3, b5 - b4 are in an A.P.

Q: Wait, why do we look at $b_{n+1} - b_n$ instead of just solving for $b_n$ directly?

That is a fantastic question! If you try to solve for $b_n$ directly, you will find yourself stuck in a recursive loop or a very complex integral that doesn't simplify easily. In JEE Advanced problems, when you see a sequence defined by an integral and options involving differences or reciprocals, it is a huge hint to look at the *difference* between consecutive terms. The structure of the problem is designed to make the integral collapse into something simple once you subtract the terms.

Q: How do I know that $\cos((2n+1)\frac{\pi}{2})$ is always zero?

Remember that $n$ is a natural number ($n \in \mathbb{N}$), so $2n+1$ is always an odd integer. Any odd integer multiplied by $\frac{\pi}{2}$ lands you on the vertical axis of the unit circle, where the cosine value is always zero. This is a classic property used in these types of problems to simplify the upper limit of integration.

Q: Why do the reciprocals form an A.P. and not the original differences?

The original differences $b_{n+1} - b_n$ are of the form $-\frac{1}{2n+1}$. As $n$ increases, these values are $-1/5, -1/7, -1/9$, etc. These are not in an A.P. because the difference between them is not constant. However, their reciprocals are $-5, -7, -9$, and the difference between these is a constant $-2$. The problem is specifically constructed to test if you can recognize this pattern in the reciprocals.

Question

If $b_n = \int_{0}^{\pi/2} \frac{\cos^2 nx}{\sin x} dx, n \in \mathbb{N}$, then

Options:

(A) $b_3 - b_2, b_4 - b_3, b_5 - b_4$ are in an A.P. with common difference $-2$

(B) $\frac{1}{b_3 - b_2}, \frac{1}{b_4 - b_3}, \frac{1}{b_5 - b_4}$ are in an A.P. with common difference $2$

(C) $b_3 - b_2, b_4 - b_3, b_5 - b_4$ are in a G.P.

(D) $\frac{1}{b_3 - b_2}, \frac{1}{b_4 - b_3}, \frac{1}{b_5 - b_4}$ are in an A.P. with common difference $-2$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

We are tasked with evaluating the integral sequence defined by:

$$b_n = \int_{0}^{\pi/2} \frac{\cos^2 nx}{\sin x} dx$$

The presence of $\sin x$ in the denominator suggests that direct integration is difficult. Instead, we focus on the relationship between consecutive terms, $b_n$ and $b_{n+1}$, to simplify the expression.

### The Strategy: Embracing the Difference

Consider the difference between consecutive terms:

$$b_{n+1} - b_n = \int_{0}^{\pi/2} \frac{\cos^2(n+1)x - \cos^2 nx}{\sin x} dx$$

To simplify the numerator, we utilize the trigonometric identity $\cos^2 A - \cos^2 B = -\sin(A+B)\sin(A-B)$. Setting $A = (n+1)x$ and $B = nx$, we find:

$$A+B = (2n+1)x \quad 	ext{and} \quad A-B = x$$

Substituting these into the identity, the numerator becomes $-\sin((2n+1)x)\sin x$.

### The Climax: The Great Cancellation

Substituting the simplified numerator back into the integral, we observe a significant simplification:

$$b_{n+1} - b_n = \int_{0}^{\pi/2} \frac{-\sin((2n+1)x) \sin x}{\sin x} dx$$

The $\sin x$ terms cancel out, leaving us with a straightforward integral:

$$b_{n+1} - b_n = -\int_{0}^{\pi/2} \sin((2n+1)x) dx$$

Evaluating this integral, we obtain:

$$b_{n+1} - b_n = -\left[ -\frac{\cos((2n+1)x)}{2n+1} ight]_{0}^{\pi/2} = \left[ \frac{\cos((2n+1)x)}{2n+1} ight]_{0}^{\pi/2}$$

### Final Calculation

At the upper limit, $\cos((2n+1)\frac{\pi}{2}) = 0$ for any integer $n$. At the lower limit, $\cos(0) = 1$. Thus:

$$b_{n+1} - b_n = \frac{0 - 1}{2n+1} = -\frac{1}{2n+1}$$

We have established the general difference formula. For specific values of $n$:

*   For $n=2$: $b_3 - b_2 = -\frac{1}{5}$
*   For $n=3$: $b_4 - b_3 = -\frac{1}{7}$
*   For $n=4$: $b_5 - b_4 = -\frac{1}{9}$

The reciprocals of these differences are $-5, -7, -9$. Since these values form an arithmetic progression with a common difference of $-2$, we conclude that the sequence of reciprocals $\frac{1}{b_{n+1} - b_n}$ forms an **A.P. with a common difference of $-2$**.

Animated Solution for Mathematics - Definite Integration: If $b_{n} = \int_{0}^{π /2} \frac{c o s ^{2} n x}{s i n x} d x, n \in N$ , then

Select Answer:

Visualized Solution

Understanding the Sequence $b_{n}$

Defining the Difference $b_{n + 1} - b_{n}$

Applying Trigonometric Identity

Simplifying the Numerator

Integrand Reduction

Integrating the Sine Function

Evaluating Limits

Calculating Specific Differences

Analyzing the Reciprocals

Identifying the A.P.

Final Conclusion

The Sigma Insight: Fundamental Theorem & Properties of Definite Integrals

Analyzing the Setup

The Strategy

The Climax

Final Calculation

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