Question

The value of the integral $\underset{0}{\overset{1}{\int }}\frac{x^3}{1+x^8}dx$ is

• A. $\frac{\pi }{16}$
• B. $\frac{\pi }{4}$
• C. $\frac{\pi }{8}$
• D. none of these

Answer 1

The correct option is A $\frac{\pi }{16}$

${\int }_0^1\frac{x^3}{1+x^8}dx$

$={\int }_0^1\frac{x^3}{1+(x^4{)}^2}dx$

$=\frac{1}{4}{\int }_0^1\frac{1}{1+(t{)}^2}dt(substitutingt=x^4\Rightarrow dt=4x^{3}dx)$

$=\frac{1}{4}{\left[{tan}^{-1}\right(t\left)\right]}_0^1$

$=\frac{1}{4}[\frac{\pi }{4}-0]$

$=\frac{\pi }{16}$