Question

If the integrals $i_1={\int }_0^{\in fty}\frac{1}{1+x^8}dx$ and $i_2={\int }_0^1\frac{1}{(1-x^8{)}^{1/8}}dx,$ then

• A. $i_1=2i_2$
• B. $i_2=8i_1$
• C. $i_1=i_2$
• D. $i_1=8i_2$

Answer 1

Answer: (C)

In $i_1$
put $x^8=ta{n}^2\theta$
$\Rightarrow 8x^{7}dx=2tan\theta se{c}^2\theta d\theta$
$\Rightarrow i_1={\int }_0^{\pi /2}\frac{2tan\theta se{c}^2\theta d\theta }{se{c}^2\theta \cdot 8\cdot ta{n}^{7/4}\theta }$
$\Rightarrow i_1=\frac{1}{4}{\int }_0^{\pi /2}ta{n}^{-3/4}\theta d\theta$
In $i_2$
put $x^8=si{n}^2\theta$
$\Rightarrow i_2={\int }_0^{\pi /2}\frac{2\text{sin }\theta \text{cos }\theta d\theta }{(1-{\left(\text{sin}\right)}^2\theta {)}^{1/8}\cdot 8{\left(\text{sin}\right)}^{7/4}\theta }$
$=\frac{1}{4}{\int }_0^{\pi /2}ta{n}^{-3/4}\theta d\theta =i_1$
$\Rightarrow i_2=i_1$