Question

If $I_n={\int }_0^1\frac{dx}{{\left(1+x^2\right)}^n};n\in N$, then which of the following statements hold good?

• A. $2n{I}_{n+1}=2^{-n}+\left(2n-1\right)I_n$
• B. $I_2=\frac{\pi }{8}+\frac{1}{4}$
• C. $I_2=\frac{\pi }{8}-\frac{1}{4}$
• D. $I_3=\frac{\pi }{16}-\frac{5}{48}$

Answer 1

Answer: (A), (B)

The correct options are
A $2n{I}_{n+1}=2^{-n}+\left(2n-1\right)I_n$
B $I_2=\frac{\pi }{8}+\frac{1}{4}$

$I_n={\int }_0^1\frac{dx}{{(1+x^2)}^n}$

$={\int }_0^{1}1\times \frac{dx}{{(1+x^2)}^n}$

Integrating by parts, we get

Let $u=\frac{1}{{(1+x^2)}^n}={(1+x^2)}^{-n}$

$\Rightarrow du=-2nx{(1+x^2)}^{-n-1}dx=\frac{-2nx}{{(1+x^2)}^{n+1}}dx$

And $dv=dx\Rightarrow v=x$

$I_n={\left[\frac{x}{{(1+x^2)}^n}\right]}_0^1-{\int }_0^{1}x\left(\frac{-2nxdx}{{(1+x^2)}^{n+1}}\right)$

$I_n=\frac{1}{2^n}+2n{\int }_0^1\left(\frac{x^{2}dx}{{(1+x^2)}^{n+1}}\right)$

$I_n=\frac{1}{2^n}+2n{\int }_0^1\left(\frac{(x^2+1-1)dx}{{(1+x^2)}^{n+1}}\right)$

$I_n=\frac{1}{2^n}+2n{\int }_0^1\left(\frac{(x^2+1)dx}{{(1+x^2)}^{n+1}}\right)-2n{\int }_0^1\left(\frac{dx}{{(1+x^2)}^{n+1}}\right)$

$I_n=\frac{1}{2^n}+2n{\int }_0^1\left(\frac{dx}{{(1+x^2)}^n}\right)-2n{\int }_0^1\left(\frac{dx}{{(1+x^2)}^{n+1}}\right)$

$\Rightarrow I_n=\frac{1}{2^n}+2n{I}_n-2n{I}_{n+1}$

$\Rightarrow 2n{I}_{n+1}=\frac{1}{2^n}+2n{I}_n-I_n$

$\Rightarrow 2n{I}_{n+1}=\frac{1}{2^n}+(2n-1)I_n$ .......$\left(1\right)$

$\Rightarrow 2n{I}_2=\frac{1}{2}+I_1$ for $n=1$

Now,$I_1={\int }_0^1\frac{dx}{1+x^2}$

$={\left[{\tan }^{-1}x\right]}_0^1$

$={\tan }^{-1}1-{\tan }^{-1}0=\frac{\pi }{4}-0=\frac{\pi }{4}$

$\therefore 2I_2=\frac{1}{2}+\frac{\pi }{4}$

$\Rightarrow I_2=\frac{1}{4}+\frac{\pi }{8}$ ..........$\left(2\right)$

From $\left(1\right)$

$4I_3=\frac{1}{2^2}+(4-1)I_2$

$\Rightarrow 4I_3=\frac{1}{2^2}+3(\frac{1}{4}+\frac{\pi }{8})$

$\Rightarrow 4I_3=\frac{1}{4}+\frac{3}{4}+\frac{3\pi }{8}=1+\frac{3\pi }{8}$

$\Rightarrow I_3=\frac{1}{4}+\frac{3\pi }{32}$

Hence option$\left(a\right)$ and option$\left(b\right)$ are correct.