Question

What is the value of \(\int_0^1 \frac{dx}{\sqrt{1+x^4}}\)?

(a) \(\beta(\frac{1}{4}, \frac{1}{2}) \)

(b) \(\frac{1}{4\sqrt{2}}\beta(\frac{1}{4},  \frac{1}{2}) \)

(c) \(\frac{1}{3\sqrt{2}}\beta(\frac{1}{3},  \frac{1}{2}) \)

(d) \(\frac{1}{4\sqrt{3}}\beta(\frac{1}{4},  \frac{1}{3}) \)

This question was posed to me in exam.

The above asked question is from Special Functions topic in chapter Special Functions – Gamma, Beta, Bessel and Legendre of Engineering Mathematics

Answer 1

Right option is (b) \(\frac{1}{4\sqrt{2}}\beta(\frac{1}{4},  \frac{1}{2}) \)

Explanation: Substitute x^2 = tan⁡θ

Therefore θ varies from 0 to \(\frac{\pi}{4}.\)

\(\int_0^{π/4} \frac{(secθ)^2}{2 sec⁡θ \sqrt{tan⁡θ}} dθ \)

\(= \int_0^{π/4}\frac{dθ}{2\sqrt{sin⁡θ cos⁡θ}} \)

\(= \frac{1}{4\sqrt{2}} \beta(\frac{1}{4},  \frac{1}{2}).\)