Question

If $I_n={\int }_0^1\frac{dx}{(1+x^2{)}^n};nϵN,$ then which of the following statements is correct ?

• A. $2n{I}_{n+1}=2^{-n}+(2n-1)I_n$
• 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

The correct option is B $I_2=\frac{\pi }{8}+\frac{1}{4}$

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

$\therefore I_2={\int }_0^1\frac{dx}{(1+x^2{)}^2}$

Put $x=\tan \theta ;dx={\sec }^2\theta d\theta$

$\therefore I_2={\int }_{a_1}^{a_2}\frac{{\sec }^2\theta d\theta }{(1+{\tan }^2\theta {)}^2}$

Also $a_2=\frac{\pi }{4}{a}_1=\mathrm{0.......}\left(\begin{array}{c}\tan \frac{\pi }{4}=1\\ \tan 0=0\end{array}\right)$

$\therefore I_2={\int }_0^{\frac{\pi }{4}}\frac{{\sec }^2\theta d\theta }{{\sec }^4\theta d\theta }$

$\therefore I_2={\int }_0^{\frac{\pi }{4}}{\cos }^2\theta d\theta$

$\therefore I_2={\int }_0^{\frac{\pi }{4}}\left(\frac{\cos 2\theta +1}{2}\right)d\theta$

$\therefore I_2=\frac{1}{2}{\int }_0^{\frac{\pi }{4}}(\cos 2\theta +1)d\theta$

$\therefore I_2=\frac{1}{2}{\left|\frac{(\sin 2\theta +\theta )}{2}\right|}_0^{\frac{\pi }{4}}$

$\therefore I_2=\frac{\pi }{8}+\frac{1}{4}$