Let ℕ be the set of natural numbers. For n∈ℕ, define I_n=∫0π(xsin^2n(x))/(sin^2n(x)+cos^2n(x))dx. Then for m,n∈ℕ

Correct answer: (C) I_m=I_n for all m≠n. I_n=1/2∫0π(xsin^2nx)/(sin^2nx+cos^2nx)+((π-x)sin^2nx)/(sin^2nx+cos^2nx)dx =(π)/2∫0π(sin^2nxdx)/(sin^2nx+cos^2nx) =2×(π)/2∫0π/2(sin^2nxdx)/(sin^2nx+cos^2nx) =(π)/2∫0π/2(sin^2nx+cos^2nx)/(sin^2nx+cos^2nx)dx =(π)/2×(π)/2=π^24 ⇒I_m=I_n∀m,n

Q53 / 91 · KVPY 2020 (SB/SX) · Mathematics

Let $ℕ$ be the set of natural numbers. For $n \in ℕ,$ define $I_{n} = \int_{0}^{π} \frac{x \sin^{2 n} (x)}{\sin^{2 n} (x) + \cos^{2 n} (x)} d x .$

Then for $m, n \in ℕ$

A
$I_{m} < I_{n}$ for all $m < n$
B
$I_{m} > I_{n}$ for all $m < n$
Correct answer
C
$I_{m} = I_{n}$ for all $m \neq n$
D
$I_{m} < I_{n}$ for some $m < n$ and $I_{m} > I_{n}$ for some $m < n$

Solution

$I_{n} = \frac{1}{2} \int_{0}^{π} (\frac{x \sin^{2 n} x}{\sin^{2 n} x + \cos^{2 n} x} + \frac{(π - x) \sin^{2 n} x}{\sin^{2 n} x + \cos^{2 n} x}) d x$

$= \frac{π}{2} \int_{0}^{π} \frac{\sin^{2 n} x d x}{\sin^{2 n} x + \cos^{2 n} x}$

$= 2 \times \frac{π}{2} \int_{0}^{π / 2} \frac{\sin^{2 n} x d x}{\sin^{2 n} x + \cos^{2 n} x}$

$= \frac{π}{2} \int_{0}^{π / 2} \frac{\sin^{2 n} x + \cos^{2 n} x}{\sin^{2 n} x + \cos^{2 n} x} d x$

$= \frac{π}{2} \times \frac{π}{2} = \frac{π^{2}}{4}$

$\Rightarrow I_{m} = I_{n} \forall m, n$

Hence, the correct answer is option: 'C'.

Let ℕ be the set of natural numbers. For n∈ℕ, define In=∫0πxsin2n(x)sin2n(x)+cos2n(x)dx. Then for m,n∈ℕ

Solution

More Definite Integration Questions

Let $ℕ$ be the set of natural numbers. For $n \in ℕ,$ define $I_{n} = \int_{0}^{π} \frac{x \sin^{2 n} (x)}{\sin^{2 n} (x) + \cos^{2 n} (x)} d x .$

Then for $m, n \in ℕ$