Question

\[ \int \frac{\sin 2 x \cos 2 x}{\sqrt{4-\cos ^4 2 x}} d x= \]

• A. \(\frac{1}{4}\mathrm{sin}^{-1}(\frac{\mathrm{cos}^{2}2x}{2})+\mathrm{c}\), where c is the constant of integration.
• B. \(\frac{-1}{4}\mathrm{sin}^{-1}(\frac{\mathrm{cos}^{2}2x}{2})+\mathrm{c}\), where c is the constant of integration.
• C. \(\frac{1}{2}\mathrm{sin}^{-1}(\frac{\mathrm{cos}^{2}2x}{2})+\mathrm{c}\), where c is the constant of integration.
• D. \(\frac{-1}{2}\mathrm{sin}^{-1}(\frac{\mathrm{cos}^{2}2x}{2})+\mathrm{c}\), where c is the constant of integration.

Hint:

Substitute u = cos^2 2x so that sin 2x cos 2x dx becomes a multiple of du.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept
The denominator has the form \(\sqrt{a^{2}-u^{2}}\) with \(u=\cos^{2}2x\) and \(a=2\).

Step 2: Substitute
\(u=\cos^{2}2x\Rightarrow du=-4\sin2x\cos2x\,dx\), so \(\sin2x\cos2x\,dx=-\dfrac{du}{4}\).

Step 3: Integrate
\(I=-\dfrac14\int\dfrac{du}{\sqrt{4-u^{2}}}=-\dfrac14\sin^{-1}\dfrac u2+c\).

Step 4: Back-substitute
\(I=-\dfrac14\sin^{-1}\left(\dfrac{\cos^{2}2x}{2}\right)+c\).

Final Answer:
The integral is -(1/4) sin^-1(cos^2(2x)/2) + c. \[ \boxed{\text{(B) }\dfrac{-1}{4}\sin^{-1}\left(\dfrac{\cos^{2}2x}{2}\right)+c} \]