hello every one prove that ∫_0 ^(π/2) cos^u (x) cos(ax) arctan(b cos(x)) dx =((2^(−u−2) .π.b.Γ(u+2))/(Γ(((u−a+3)/2))Γ(((u+a+3)/2)))).x_4 F_3 ((((1/2),1+(u/2),((u+3)/2),−b^2 )),(((3/2),((u−a+3)/2),((u+a+3)/2))) ) Re u>−1 ,∣arg(1+b^2 ) ∣<π


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Question Number 100829 by M±th+et+s last updated on 28/Jun/20

$h e l l o e v e r y o n e$ $p r o v e t h a t$ $\int_{0}^{\frac{π}{2}} {c o s}^{u} (x) c o s (a x) a r c t a n (b c o s (x)) d x$ $= \frac{2^{- u - 2} . π . b . Γ (u + 2)}{Γ (\frac{u - a + 3}{2}) Γ (\frac{u + a + 3}{2})} . x_{4} F_{3} (\begin{matrix} \frac{1}{2}, 1 + \frac{u}{2}, \frac{u + 3}{2}, - b^{2} \\ \frac{3}{2}, \frac{u - a + 3}{2}, \frac{u + a + 3}{2} \end{matrix})$ $R e u > - 1, ∣ a r g (1 + b^{2}) ∣< π$
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