Key Concepts and Formulas

• Trigonometric Identity: $\sin 2x = 2 \sin x \cos x$
• Trigonometric Identity: $\sec^2 x = 1 + \tan^2 x$
• Substitution Method for Integration: If $\int f(g(x))g'(x) \, dx$ is to be evaluated, substitute $t = g(x)$, so $dt = g'(x) \, dx$.

Step-by-Step Solution

Step 1: Rewrite the integral using the double angle formula.

We are given the integral $\int \frac{dx}{\cos^3 x \sqrt{2 \sin 2x}}$. We can rewrite $\sin 2x$ as $2 \sin x \cos x$, so the integral becomes:

$\int \frac{dx}{\cos^3 x \sqrt{2(2 \sin x \cos x)}} = \int \frac{dx}{\cos^3 x \sqrt{4 \sin x \cos x}}$

Step 2: Simplify the expression.

We can simplify the square root:

$\int \frac{dx}{\cos^3 x (2 \sqrt{\sin x \cos x})} = \int \frac{dx}{2 \cos^3 x \sqrt{\sin x \cos x}} = \int \frac{dx}{2 \cos^4 x \sqrt{\frac{\sin x}{\cos x}}}$

Step 3: Express the integral in terms of $\tan x$.

We know that $\frac{\sin x}{\cos x} = \tan x$. Thus, we can rewrite the integral as:

$\int \frac{dx}{2 \cos^4 x \sqrt{\tan x}}$

Also, $\frac{1}{\cos^2 x} = \sec^2 x$. Then $\frac{1}{\cos^4 x} = \sec^4 x$. So,

$\int \frac{\sec^4 x}{2 \sqrt{\tan x}} dx = \frac{1}{2} \int \frac{\sec^2 x \cdot \sec^2 x}{\sqrt{\tan x}} dx$

Step 4: Use the identity $\sec^2 x = 1 + \tan^2 x$ to rewrite the integral.

$\frac{1}{2} \int \frac{(1 + \tan^2 x) \sec^2 x}{\sqrt{\tan x}} dx$

Step 5: Perform a substitution.

Let $t = \tan x$. Then $dt = \sec^2 x \, dx$. Also, $\tan x = t$, so $\sqrt{\tan x} = \sqrt{t} = t^{1/2}$. The integral becomes:

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

Step 6: Evaluate the integral.

$\frac{1}{2} \left[ \frac{t^{1/2}}{1/2} + \frac{t^{5/2}}{5/2} \right] + k = \frac{1}{2} \left[ 2t^{1/2} + \frac{2}{5} t^{5/2} \right] + k = t^{1/2} + \frac{1}{5} t^{5/2} + k$

Step 7: Substitute back to get the integral in terms of $x$.

Since $t = \tan x$, we have:

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

Step 8: Compare with the given expression.

We are given that the integral is equal to $(\tan x)^A + C (\tan x)^B + k$. Comparing this with our result $(\tan x)^{1/2} + \frac{1}{5} (\tan x)^{5/2} + k$, we have $A = \frac{1}{2}$, $B = \frac{5}{2}$, and $C = \frac{1}{5}$.

Step 9: Calculate A + B + C.

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

Common Mistakes & Tips

• Remember the trigonometric identities, especially those involving $\sin 2x$ and $\sec^2 x$. These are frequently used in integration problems.
• Be careful when substituting back after integration. Ensure you substitute correctly.
• Double check your arithmetic when adding fractions.

Summary

We evaluated the integral $\int \frac{dx}{\cos^3 x \sqrt{2 \sin 2x}}$ by using trigonometric identities to rewrite the integral in terms of $\tan x$. We then used a substitution to simplify the integral and evaluated it. Finally, we compared the result with the given expression to find the values of A, B, and C, and calculated their sum. The final answer is $\frac{16}{5}$.

Final Answer The final answer is \boxed{\frac{16}{5}}, which corresponds to option (B).