advanced-calculus-prove-that-0-pi-3-dx-cos-2-x-1-3-2-1-3-pi-3-

Question Number 130844 by mnjuly1970 last updated on 29/Jan/21

\dots a d v a n c e d c a l c u l u s \dots

p r o v e t h a t ::

\int_{0}^{\frac{π}{3}} \frac{d x}{\sqrt[3]{{c o s}^{2} (x)}} = \frac{2^{\frac{1}{3}} \sqrt{π}}{\sqrt{3}}

Answered by Dwaipayan Shikari last updated on 29/Jan/21

\int_{0}^{\frac{π}{3}} {(c o s x)}^{- \frac{2}{3}} d x = - \int_{1}^{\frac{1}{2}} t^{- \frac{2}{3}} {(1 - t^{2})}^{- \frac{1}{2}} d t

= \int_{0}^{1} t^{- \frac{2}{3}} {(1 - t^{2})}^{- \frac{1}{2}} d t - \int_{0}^{\frac{1}{2}} t^{- \frac{5}{6}} {(1 - t^{2})}^{- \frac{1}{2}} d t

= \frac{1}{2} \int_{0}^{1} u^{- \frac{5}{6}} {(1 - u)}^{- \frac{1}{2}} d u - \frac{1}{2} \int_{0}^{\frac{1}{\sqrt{2}}} u^{- \frac{5}{6}} {(1 - u)}^{- \frac{1}{2}} d u

= \frac{Γ (\frac{1}{6}) Γ (\frac{1}{2})}{2 Γ (\frac{2}{3})} - \frac{1}{2} \underset{n = 0}{\sum^{\infty}} \frac{{(\frac{1}{2})}_{n}}{n!} \int_{0}^{\frac{1}{\sqrt{2}}} u^{n - \frac{5}{6}} d u

= \frac{Γ (\frac{5}{6}) \sqrt{π}}{2 Γ (\frac{2}{3})} - \frac{3}{\sqrt[3]{2}}_{2} F_{1} (\frac{1}{6}, \frac{1}{2}, \frac{7}{6}, \frac{1}{4})