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: Rewrite $\cos 2x$ to expose a tangent.
Use $\cos 2x = \cos^2 x - \sin^2 x = \cos^2 x(1 - \tan^2 x)$.
Step 2: Substitute into the radical.
$\sqrt{\cos 2x} = \cos x\,\sqrt{1 - \tan^2 x}$ (taking $\cos x > 0$ on the working interval).
Step 3: Put this into the integral.
$\displaystyle\int \dfrac{dx}{\cos x\cdot\cos x\sqrt{1 - \tan^2 x}} = \int \dfrac{dx}{\cos^2 x\,\sqrt{1 - \tan^2 x}}$.
Step 4: Recognise $\sec^2 x$.
$\dfrac{1}{\cos^2 x} = \sec^2 x$, so the integral is $\displaystyle\int \dfrac{\sec^2 x}{\sqrt{1 - \tan^2 x}}\,dx$.
Step 5: Substitute $t = \tan x$.
Then $dt = \sec^2 x\,dx$, turning the integral into $\displaystyle\int \dfrac{dt}{\sqrt{1 - t^2}}$.
Step 6: Integrate and back-substitute.
$\displaystyle\int \dfrac{dt}{\sqrt{1 - t^2}} = \sin^{-1}t + c = \sin^{-1}(\tan x) + c$.
\[ \boxed{\sin^{-1}(\tan x) + c} \]