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From the first step written in my previous post, I proceed further,

Put \(\csc{(x)}=\frac{\sec{(x)}}{\tan{(x)}}\) and \(\tan^2{(x)}=\sec^2{(x)}-1\)

\(\displaystyle\int\sqrt{4*\csc^2{(x)}+1}dx=\displaystyle\int\frac{\sqrt{5*\sec^2{(x)}-1}*\sec{(x)}\tan{(x)}}{\sec{(x)}*(\sec^2{(x)}-1)}dx ……... {(1)}\)

Substitute \(u=\sec{(x)}\), the L.H.S. of (1)=\(\displaystyle\int\frac{\sqrt{5*u^2-1}}{u*(u^2-1)}du\)

Substitute\(v=\sqrt{5*u^2-1}\), the L.H.S of (1) =\(5*\displaystyle\int\frac{v^2}{(v^2-4)*(v+1)}dv\)

So, L.H.S. of (1)=\(-\ln{(v+2)}+\arctan{(v)}+\ln{(v-2)}\)

Undoing the substitution \(v =\sqrt{5*u^2-1}\) and \(u=\sec{(x)}\)

The L.H.S.of (1)=\(-\ln{(\sqrt{5*\sec^2{(x)}-1}+2)}+\arctan{(\sqrt{(5*\sec^2{(x)}-1)})} + \ln{(\sqrt{5*\sec^2{(x)}-1}-2)}\)

Hence, the definite integral of the L.H.S.of (1) from \(\frac{\pi}{4}\) to \(\frac{\pi}{2} = \frac{\pi}{2} +\ln{(5)}-\arctan{(3)}\)

Hence

** A+B+C=10, where A=2, B=5 and C=3. (QED)**