Gamma Integral

$\int_0^{π/2} \frac{1}{\sqrt{8\cos^4 x +7\sin^4 x}} dx$

$\int_0^{π/2} \frac{sec^2 x }{\sqrt{8 + 7tan^4 x }} dx$

$t= tanx$

$\int_0^∞ \frac{dt}{\sqrt{8+7 t^4}}$
$8+7 t^4 = u^2 , 2.7.t^3= u \frac{du}{dt}$ Integral becomes $\frac{1}{2\sqrt[4]{7}} \int_0^∞ \frac{du}{(u^2 -8 )^{\frac{3}{4}}}$

$u^2 -8 = p , 2u=\frac{dp}{du}$ $\frac{1}{4\sqrt[4]{8^2 7}} \int_0^∞ \frac{dp}{\sqrt{(1+\frac{p}{8} ) (p^{\frac{3}{2}})}}$ $\frac{p}{8} =ü$ Integral becomes $\frac{1}{4\sqrt[4]{56}} \int_0^∞ \frac{ ü^{-3/4} dü}{(\sqrt{(1+ü)})}$

Take $\frac{ü}{1+ü} = μ$ After simplifying the integral becomes $\frac{1}{4\sqrt[4]{56}} \int_0^1 (1-μ)^{-3/4} μ^{-3/4} dμ$

Which is in a closed form of $\frac{1}{4\sqrt[4]{56}} β(\frac{1}{4} , \frac{1}{4} )$

Which can also be written as $\frac{1}{4\sqrt[4]{56}} \frac{Γ^2 (\frac{1}{4})}{Γ(\frac{1}{2})}$

Or other form $\frac{1}{4\sqrt[4]{π^2 56}} Γ^2 (\frac{1}{4})$ as $Γ(1/2) = √π$

As per question $a = 4 , b= 4 , c=2$

Final answer $\boxed{(a-b)c =0}🍎🍏$