\(\int_{0}^{1}\displaystyle\frac{x^{3}}{1+x^{8}}\,dx\) is equal to
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \(\displaystyle\frac{\pi}{16}\)
Since \(x^{8} = (x^{4})^{2}\), let \(t = x^{4}\), so \(dt = 4x^{3}\,dx\).
When \(x = 0\), \(t = 0\); when \(x = 1\), \(t = 1\). Then
\( \displaystyle \int_{0}^{1}\frac{x^{3}}{1+x^{8}}\,dx = \frac{1}{4}\int_{0}^{1}\frac{dt}{1+t^{2}} \)
\( \displaystyle = \frac{1}{4}\left[\tan^{-1}t\right]_{0}^{1} \)
\( \displaystyle = \frac{1}{4}\cdot\frac{\pi}{4} = \frac{\pi}{16} \)
Why the other options are incorrect
Option (1) \(\displaystyle\frac{\pi}{4}\) — Incorrect. It forgets the factor \(\frac{1}{4}\) from \(x^{3}\,dx = \frac{1}{4}\,dt\).
Option (2) \(\displaystyle\frac{\pi}{8}\) — Incorrect. It uses \(\frac{1}{2}\) instead of \(\frac{1}{4}\), as if t were \(x^{2}\).
Option (3) \(\displaystyle\frac{\pi}{32}\) — Incorrect. It uses \(\frac{1}{8}\), taking the power 8 instead of the power 4 in the substitution.