Question

Evaluate the indefinite integral $$\int \frac{dx}{\cos x \sqrt{\cos 2x}}$$

• A. $\sin^{-1}(\tan x) + c$
• B. $\frac{1}{2} \log\left| \tan\left(\frac{\pi}{4} + x\right) \right| + c$
• C. $2 \log\left| \frac{1 + \tan x}{1 - \tan x} \right| + c$
• D. $\frac{1}{2} \log\left| \frac{1 - \tan x}{1 + \tan x} \right| + c$

Hint:

Whenever you see $\sqrt{\cos 2x}$ in a denominator, factoring out $\cos^2 x$ from under the radical is a classic integration trick! It instantly generates a $\sec^2 x$ in the numerator and leaves a clean $(1 - \tan^2 x)$ inside the root, setting up a perfect substitution path.

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
The problem requires us to find the antiderivative of an expression containing a product of a cosine function and a square root of a double-angle cosine function in the denominator.

Step 2: Key Formula or Approach:
We will convert the double-angle cosine term inside the radical using a standard trigonometric identity:
$$\cos 2x = \frac{1 - \tan^2 x}{1 + \tan^2 x}$$ Alternatively, we can express it in terms of cosine and sine components ($\cos 2x = \cos^2 x - \sin^2 x$) and factor out a $\cos x$ term to create secant and tangent structures.

Step 3: Detailed Explanation:
Let's modify the interior of the square root radical:
$$\cos 2x = \cos^2 x - \sin^2 x = \cos^2 x \left(1 - \frac{\sin^2 x}{\cos^2 x}\right) = \cos^2 x (1 - \tan^2 x)$$ Substitute this identity back into the main integral expression:
$$I = \int \frac{dx}{\cos x \sqrt{\cos^2 x (1 - \tan^2 x)}}$$ Pull the $\cos^2 x$ out from under the radical square root sign as $\cos x$:
$$I = \int \frac{dx}{\cos x \cdot \cos x \sqrt{1 - \tan^2 x}} = \int \frac{dx}{\cos^2 x \sqrt{1 - \tan^2 x}}$$ We know that $\frac{1}{\cos^2 x} = \sec^2 x$, so the integral becomes:
$$I = \int \frac{\sec^2 x}{\sqrt{1 - \tan^2 x}} \, dx$$ Now, perform a variable substitution by setting $t = \tan x$. Differentiating both sides gives:
$$dt = \sec^2 x \, dx$$ Substitute these into our rearranged integral:
$$I = \int \frac{dt}{\sqrt{1 - t^2}}$$ This matches the standard integration form $\int \frac{1}{\sqrt{1-t^2}} \, dt = \sin^{-1} t + c$. Evaluating the integral gives:
$$I = \sin^{-1}(t) + c$$ Substitute back $t = \tan x$ to return to our original variable:
$$I = \sin^{-1}(\tan x) + c$$ This matches option (A).

Step 4: Final Answer:
The value of the indefinite integral is $\sin^{-1}(\tan x) + c$, which corresponds to option (A).