Q80 / 150 · MHT CET 2025 (22 Apr Shift 1) · Mathematics
$\int \frac{\sin 2 x \cos 2 x}{\sqrt{4-\cos ^4 2 x}} \mathrm{~d} x=$
- A$\frac{1}{4} \sin ^{-1}\left(\frac{\cos ^2 2 x}{2}\right)+c \quad$, where $c$ is the constant of integration.
- Correct answerB$\frac{-1}{4} \sin ^{-1}\left(\frac{\cos ^2 2 x}{2}\right)+c$, where $c$ is the constant of integration.
- C$\frac{1}{2} \sin ^{-1}\left(\frac{\cos ^2 2 x}{2}\right)+c \quad$, where $c$ is the constant of integration.
- D$\frac{-1}{2} \sin ^{-1}\left(\frac{\cos ^2 2 x}{2}\right)+c \quad$, where $c$ is the constant of integration.
Solution
Let $I = \int \frac{\sin 2x \cos 2x}{\sqrt{4 - \cos^4 2x}} dx$.
Applying the substitution $u = \cos^2 2x$, the differential becomes $du = -4 \sin 2x \cos 2x dx$, yielding $\sin 2x \cos 2x dx = -\frac{1}{4} du$.
Substituting into the integral gives $I = -\frac{1}{4} \int \frac{1}{\sqrt{4 - u^2}} du$.
This evaluates to the inverse sine form: $I = -\frac{1}{4} \sin^{-1}\left(\frac{u}{2}\right) + c$.
Returning to the original variable, $I = -\frac{1}{4} \sin^{-1}\left(\frac{\cos^2 2x}{2}\right) + c$.
Final Answer: $\boxed{\text{B}}$
Hence, the correct answer is option: 'B'.