Question

The value of the integral $\displaystyle \int_0^1$ $\frac{x^{3}}{1+x^{8}} dx$ is equal to

• A. $\frac{\pi}{8}$
• B. $\frac{\pi}{4}$
• C. $\frac{\pi}{16}$
• D. $\frac{\pi}{6}$

Answer 1

The Correct Option is C

Let $I=\displaystyle\int_{0}^{1} \frac{x^{3}}{1+x^{8}} d x$
Put $x^{4}=t $
$ \Rightarrow 4 x^{3}\, d x=d t$
$\therefore I=\displaystyle \int_{0}^{1} \frac{1}{1+t^{2}} \cdot \frac{1}{4} d t=\frac{1}{4}\left[\tan ^{-1} t\right]_{0}^{1}$
$=\frac{1}{4}\left[\tan ^{-1} 1-\tan ^{-1} 0\right]=\frac{1}{4}\left[\frac{\pi}{4}-0\right]=\frac{\pi}{16}$