If $$\int \frac{dx}{\cos^3 x \sqrt{2\sin 2x}} = (\tan x)^A + C(\tan x)^B + k$$, where k is a constant of integration, then $$A + B + C$$ equals

We start with the integral:

$$\int \frac{dx}{\cos^3 x \sqrt{2\sin 2x}}$$

Recall that $$\sin 2x = 2 \sin x \cos x$$, so:

$$\sqrt{2\sin 2x} = \sqrt{2 \cdot 2 \sin x \cos x} = \sqrt{4 \sin x \cos x} = 2 \sqrt{\sin x \cos x}$$

Substituting this into the integral:

$$\int \frac{dx}{\cos^3 x \cdot 2 \sqrt{\sin x \cos x}} = \frac{1}{2} \int \frac{dx}{\cos^3 x \sqrt{\sin x \cos x}}$$

Simplify the denominator:

$$\cos^3 x \sqrt{\sin x \cos x} = \cos^3 x \cdot (\sin x \cos x)^{1/2} = \cos^3 x \cdot \sin^{1/2} x \cos^{1/2} x = \sin^{1/2} x \cos^{7/2} x$$

So the integral becomes:

$$\frac{1}{2} \int \frac{dx}{\sin^{1/2} x \cos^{7/2} x} = \frac{1}{2} \int \sin^{-1/2} x \cos^{-7/2} x dx$$

Use the substitution $$t = \tan x$$. Then $$dt = \sec^2 x dx = (1 + \tan^2 x) dx = (1 + t^2) dx$$, so $$dx = \frac{dt}{1+t^2}$$.

Express $$\sin x$$ and $$\cos x$$ in terms of $$t$$:

$$\sin x = \frac{t}{\sqrt{1+t^2}}, \quad \cos x = \frac{1}{\sqrt{1+t^2}}$$

Substitute into the integral:

$$\sin^{-1/2} x \cos^{-7/2} x = \left( \frac{t}{\sqrt{1+t^2}} \right)^{-1/2} \left( \frac{1}{\sqrt{1+t^2}} \right)^{-7/2} = t^{-1/2} (1+t^2)^{1/4} \cdot (1+t^2)^{7/4} = t^{-1/2} (1+t^2)^{(1/4 + 7/4)} = t^{-1/2} (1+t^2)^2$$

Now the integral is:

$$\frac{1}{2} \int t^{-1/2} (1+t^2)^2 \cdot \frac{dt}{1+t^2} = \frac{1}{2} \int t^{-1/2} (1+t^2) dt$$

Simplify:

$$\frac{1}{2} \int \left( t^{-1/2} + t^{3/2} \right) dt$$

Integrate term by term:

$$\frac{1}{2} \int t^{-1/2} dt + \frac{1}{2} \int t^{3/2} dt = \frac{1}{2} \cdot \frac{t^{1/2}}{1/2} + \frac{1}{2} \cdot \frac{t^{5/2}}{5/2} + k = \frac{1}{2} \cdot 2 t^{1/2} + \frac{1}{2} \cdot \frac{2}{5} t^{5/2} + k = t^{1/2} + \frac{1}{5} t^{5/2} + k$$

Substitute back $$t = \tan x$$:

$$\sqrt{\tan x} + \frac{1}{5} (\tan x)^{5/2} + k = (\tan x)^{1/2} + \frac{1}{5} (\tan x)^{5/2} + k$$

Compare to the given form $$(\tan x)^A + C(\tan x)^B + k$$:

So, $$A = \frac{1}{2}$$, $$B = \frac{5}{2}$$, and $$C = \frac{1}{5}$$.

Now compute $$A + B + C$$:

$$\frac{1}{2} + \frac{5}{2} + \frac{1}{5} = \frac{6}{2} + \frac{1}{5} = 3 + \frac{1}{5} = \frac{15}{5} + \frac{1}{5} = \frac{16}{5}$$

Hence, the correct answer is Option A.