If $I_n = \int_0^{\pi/4} \tan^n x dx$ (for $n > 1$), then $I_n + I_{n-2}$ equals:

1. Combine Integrals: $I_n + I_{n-2} = \int_0^{\pi/4} (\tan^n x + \tan^{n-2} x) dx$ $= \int_0^{\pi/4} \tan^{n-2} x (\tan^2 x + 1) dx$ $= \int_0^{\pi/4} \tan^{n-2} x \sec^2 x dx$. 2. Substitution: Let $u = \tan x$, then $du = \sec^2 x dx$. Limits: $x=0 \implies u=0; x=\pi/4 \implies u=1$. 3. Integrate: $\int_0^1 u^{n-2} du = [\frac{u^{n-1}}{n-1}]_0^1 = \frac{1}{n-1}$.

Definite Integrals

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If $I_{n} = \int_{0}^{π /4} tan^{n} x d x$ (for $n > 1$ ), then $I_{n} + I_{n - 2}$ equals:

Select the correct option:

$1/ n$

$1/ (n - 1)$

$1/ (n + 1)$

$2/ n$

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About This Question

Subject: mathematics
Chapter: integral calculus
Topic: definite integrals
Difficulty: Medium
Year: 2025

Solution

Correct Answer:

$1/ (n - 1)$

Combine Integrals: $I_{n} + I_{n - 2} = \int_{0}^{π /4} (tan^{n} x + tan^{n - 2} x) d x$ $= \int_{0}^{π /4} tan^{n - 2} x (tan^{2} x + 1) d x$ $= \int_{0}^{π /4} tan^{n - 2} x sec^{2} x d x$ .
Substitution: Let $u = tan x$ , then $d u = sec^{2} x d x$ . Limits: $x = 0 ⟹ u = 0; x = π /4 ⟹ u = 1$ .
Integrate: $\int_{0}^{1} u^{n - 2} d u = [\frac{u ^{n - 1}}{n - 1}]_{0}^{1} = \frac{1}{n - 1}$ .

This medium difficulty mathematics question is from the chapter integral calculus, covering the topic of definite integrals. It appeared in the 2025 exam.

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