If I n -01 dx -1- x 2 n- where n - N- then which of the following is-are correct-A- 2 nI n -1-2- n -2 n -1 I nI 2-1-4-8C- I 2-8-1-4D- I 3-1-4-3 -32

The correct options are A 2nI_n+1=2^ n+(2n 1)I_n nbsp; nbsp; B I_2=14+π8 nbsp; D I_3=14+3π32 nbsp;I_n=∫_0^1dx(1+x^2)^n =∫_0^1(1+x^2)^ n dx =x(1+x^2)^n|_ ...

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Byju's Answer

Standard XII

Mathematics

Summation Using Sigma

If I n =∫01 d...

Question

If $I_{n} = 1 \int 0 \frac{d x}{(1 + x^{2})^{n}}$ , where $n \in N$ , then which of the following is/are correct ?

2 n I_{n + 1} = 2^{- n} + (2 n - 1) I_{n}

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I_{2} = \frac{1}{4} + \frac{π}{8}

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I_{2} = \frac{π}{8} - \frac{1}{4}

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I_{3} = \frac{1}{4} + \frac{3 π}{32}

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Solution

The correct options are
A $2 n I_{n + 1} = 2^{- n} + (2 n - 1) I_{n}$
B $I_{2} = \frac{1}{4} + \frac{π}{8}$
D $I_{3} = \frac{1}{4} + \frac{3 π}{32}$
$I_{n} = 1 \int 0 \frac{d x}{(1 + x^{2})^{n}}$
$= 1 \int 0 (1 + x^{2})^{- n} d x$
$= \frac{x}{(1 + x^{2})^{n}} {∣ ∣ ∣}_{0}^{1} - 1 \int 0 (- n) (1 + x^{2})^{- n - 1} \cdot 2 x \cdot x d x$
$= \frac{1}{2^{n}} + 2 n 1 \int 0 \frac{x^{2} d x}{(1 + x^{2})^{n + 1}}$
$= \frac{1}{2^{n}} + 2 n 1 \int 0 \frac{1 + x^{2} - 1}{(1 + x^{2})^{n + 1}} d x$
$\Rightarrow I_{n} = \frac{1}{2^{n}} + 2 n I_{n} - 2 n I_{n + 1}$
$\Rightarrow 2 n I_{n + 1} = 2^{- n} + (2 n - 1) I_{n}$

Put $n = 1$ , we get
$2 I_{2} = \frac{1}{2} + I_{1} = \frac{1}{2} + [{tan}^{- 1} x]_{0}^{1}$
$\Rightarrow I_{2} = \frac{1}{4} + \frac{π}{8}$

$4 I_{3} = 2^{- 2} + 3 I_{2}$
$= \frac{1}{4} + 3 (\frac{1}{4} + \frac{π}{8})$
$= \frac{1}{4} + \frac{3 π}{32}$

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