If \(\int \frac{\mathrm{d} x}{\cos ^3 x \sqrt{2 \sin 2 x}}=(\tan x)^{\mathrm{A}}+\mathrm{C}(\tan x)^{\mathrm{B}}+\mathrm{K}\), where K is a constant of integration, then the value of \(5(\mathrm{~A}+\mathrm{B}+\mathrm{C})\) is equal to
[Note: The question has been modified to get the correct answer.]

\(\begin{aligned}
& \text { Let } \mathrm{I}=\int \frac{\mathrm{d} x}{\cos ^3 x \sqrt{2 \sin 2 x}} \\
& \text { Put } \sin 2 x=\frac{2 \tan x}{1+\tan ^2 x} \\
& \therefore \quad \mathrm{I}=\int \frac{\sec ^2 x \cdot \sec ^2 x}{2 \sqrt{x \tan x}} \mathrm{~d} x
\end{aligned}\)
Let \(\tan x=\mathrm{t}\)
\(\therefore \quad \sec ^2 \mathrm{~d} x=\mathrm{dt}\)
\(\begin{aligned} \therefore \quad I & =\frac{1}{2} \int \frac{1+t^2}{\sqrt{t}} d t=\frac{1}{2} \int\left(\frac{1}{\sqrt{t}}+t^{\frac{3}{2}}\right) d t \\ & =\frac{1}{2}\left[\frac{t^{\frac{1}{2}}}{\frac{1}{2}}+\frac{t^{\frac{5}{2}}}{\frac{5}{2}}\right]+k=(\tan \theta)^{\frac{1}{2}}+\frac{1}{5}(\tan \theta)^{\frac{5}{2}}+k\end{aligned}\)
Comparing with \((\tan x)^{\mathrm{A}}+\mathrm{C}(\tan x)^{\mathrm{B}}+\mathrm{k}\), we get
\(\begin{aligned}
& A=\frac{1}{2}, B=\frac{5}{2}, C=\frac{1}{5} \\
\therefore \quad & 5(A+B+C)=5\left(\frac{1}{2}+\frac{5}{2}+\frac{1}{5}\right)=16
\end{aligned}\)
[Note: In the question, \(\int \frac{\mathrm{d} x}{\cos ^3 x \sqrt{x \sin 2 x}}\) is changed to \(\int \frac{\mathrm{d} x}{\cos ^3 x \sqrt{2 \sin 2 x}}\) to apply appropriate textual concepts.]