Question

If the integrals $i_{1}=\displaystyle \int _{0}^{\in fty} \frac{1}{1 + x^{8}}dx$ and $i_{2}=\displaystyle \int _{0}^{1} \frac{1}{\left(\right. 1 - x^{8} \left.\right)^{1 / 8}}dx,$ 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}=tan^{2}\theta $
$\Rightarrow 8x^{7}dx=2tan \theta \, s e c^{2}\theta d\theta $
$\Rightarrow i_{1}=\displaystyle \int _{0}^{\pi / 2} \frac{2 tan \theta \, s e c^{2} \theta d \theta }{s e c^{2} \theta \cdot 8 \cdot t a n^{7 / 4} \theta }$
$\Rightarrow i_{1}=\frac{1}{4}\displaystyle \int _{0}^{\pi / 2} t a n^{- 3 / 4}\theta d\theta $
In $i_{2}$
put $x^{8}=s i n^{2}\theta $
$\Rightarrow i_{2}=\displaystyle \int _{0}^{\pi / 2} \frac{2 \text{sin } \text{} \theta \, \text{cos } \theta d \theta }{\left(\right. 1 - \left(\text{sin}\right)^{2} \theta \left.\right)^{1 / 8} \cdot 8 \left(\text{sin}\right)^{7/4} \theta }$
$=\frac{1}{4}\displaystyle \int _{0}^{\pi / 2} t a n^{- 3 / 4}\theta d\theta =i_{1}$
$\Rightarrow i_{2}=i_{1}$