Iₙ = ∫₀π/⁴ tanⁿ x dx then limₙ→∞ n[Iₙ + Iₙ₊₂] equals
(a) 1/2
(b) 1
(c) ∞
(d) zero

Q: Why can't we just find the value of $I_n$ using a reduction formula and then take the limit?

You absolutely could! However, using the reduction formula $I_n + I_{n+2} = \frac{1}{n+1}$ is essentially what we are doing here, just in a more direct, elegant way. Finding the general form of $I_n$ is often tedious and prone to algebraic errors; combining the terms first is the 'pro' move that saves time and reduces complexity.

Q: Wait, why does the limit of $n/(n+1)$ become 1? Shouldn't it be 0 because the denominator is larger?

That is a common trap! Remember that as $n$ grows to infinity, the difference between $n$ and $n+1$ becomes negligible compared to their magnitude. Mathematically, we divide both by $n$ to get $1 / (1 + 1/n)$. As $n \to \infty$, the $1/n$ term vanishes to zero, leaving us with $1/1$, which is 1.

Q: Is it always valid to combine integrals like this?

Yes, provided the limits of integration are identical! Since both $I_n$ and $I_{n+2}$ are defined over the same interval $[0, \pi/4]$, the linearity of the integral allows us to add the integrands directly. Always check your limits first!

Q: What happens if $n$ is not an integer? Does the logic still hold?

The logic holds for any real $n > -1$. The substitution $u = \tan x$ works perfectly fine for non-integer powers. The beauty of this method is its robustness; it doesn't rely on the discrete nature of $n$.

Q: I feel like I'm missing something about the $\tan^n x$ behavior. Does it matter that $x$ is only up to $\pi/4$?

It matters immensely! Because $x$ is restricted to $[0, \pi/4]$, the value of $\tan x$ is always between 0 and 1. This ensures that $\tan^n x$ behaves nicely as $n$ increases, shrinking toward zero. This convergence is exactly why the limit exists and is finite.

Question

$I_n = \int_{0}^{\pi/4} 	an^n x dx$ then $\lim_{n 	o \infty} n[I_n + I_{n+2}]$ equals

Options:

(A) 1/2

(B) 1

(C) $\infty$

(D) zero

Sigma Percentile · Accepted Answer

### Analyzing the Setup

We are tasked with evaluating the limit of a sequence of integrals: $\lim_{n 	o \infty} n[I_n + I_{n+2}]$, where $I_n = \int_{0}^{\pi/4} 	an^n x dx$.

Many students see the $n$ and the integral and immediately reach for complex reduction formulas. However, in JEE Advanced, the most elegant path is rarely the longest one. Let us pause and look at the structure of our expression.

### The Power of Synthesis

We are given $I_n = \int_{0}^{\pi/4} 	an^n x dx$. Our target is the sum $I_n + I_{n+2}$. By the linearity of the integral, we can write:

$$I_n + I_{n+2} = \int_{0}^{\pi/4} 	an^n x dx + \int_{0}^{\pi/4} 	an^{n+2} x dx = \int_{0}^{\pi/4} (	an^n x + 	an^{n+2} x) dx$$

We observe a common factor of $	an^n x$ waiting to be pulled out. Factoring this term yields:

$$\int_{0}^{\pi/4} 	an^n x (1 + 	an^2 x) dx$$

### The Trigonometric Key

Recall the fundamental trigonometric identity $1 + 	an^2 x = \sec^2 x$. Substituting this into our integral, the expression transforms into:

$$\int_{0}^{\pi/4} 	an^n x \sec^2 x dx$$

This is a classic setup for the substitution method. Since the derivative of $	an x$ is $\sec^2 x$, we set $t = 	an x$, which implies $dt = \sec^2 x dx$.

### The Transformation

We must update our limits of integration accordingly. When $x = 0$, $t = 	an(0) = 0$. When $x = \pi/4$, $t = 	an(\pi/4) = 1$.

Our integral now simplifies to a basic power rule problem:

$$I_n + I_{n+2} = \int_{0}^{1} t^n dt$$

Evaluating this integral is straightforward. Applying the power rule and the limits from $0$ to $1$, we obtain:

$$\left[ \frac{t^{n+1}}{n+1} ight]_{0}^{1} = \frac{1^{n+1}}{n+1} - 0 = \frac{1}{n+1}$$

### The Final Ascent

We are now ready to find the limit as $n 	o \infty$ of $n[I_n + I_{n+2}]$. Substituting our result, we have:

$$\lim_{n 	o \infty} n \cdot \frac{1}{n+1} = \lim_{n 	o \infty} \frac{n}{n+1}$$

To evaluate this limit, we divide the numerator and the denominator by $n$:

$$\lim_{n 	o \infty} \frac{1}{1 + \frac{1}{n}}$$

As $n$ approaches infinity, the term $\frac{1}{n}$ vanishes to zero. We are left with $\frac{1}{1+0}$, which results in the final answer of **1**.

Animated Solution for Mathematics - Definite Integration: $I_{n} = \int_{0}^{π /4} tan^{n} x d x$ then $lim_{n \to \infty} n [I_{n} + I_{n + 2}]$ equals

Select Answer:

Visualized Solution

Understanding $I_{n}$

The Target Expression

Combining the Integrals

Factoring the Integrand

Trigonometric Identity

The Substitution Method

Changing the Limits

The Transformed Integral

Evaluating the Integral

Setting up the Limit

Evaluating the Limit

Final Conclusion

The Sigma Insight: Newton-Leibniz & Reduction Formulas

Analyzing the Setup

The Power of Synthesis

The Trigonometric Key

The Transformation

The Final Ascent

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Animated Solution for Mathematics - Definite Integration: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } In​=∫0π/4​tannxdx then limn→∞​n[In​+In+2​] equals