\(\int\displaystyle\frac{\sin 2x\cos 2x}{\sqrt{9-\cos^4 2x}}\,dx\) is equal to
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \(\displaystyle -\frac{1}{4}\sin^{-1}\left(\frac{1}{3}\cos^2 2x\right)+C\)
Since \(\cos^4 2x = \left(\cos^2 2x\right)^2\), the substitution \(t = \cos^2 2x\) turns the denominator into \(\sqrt{9-t^2}\). By the chain rule:
\( dt = 2\cos 2x\times(-2\sin 2x)\,dx \)
\( \displaystyle \sin 2x\cos 2x\,dx = -\frac{dt}{4} \)
Then
\( \displaystyle \int\frac{\sin 2x\cos 2x}{\sqrt{9-\cos^4 2x}}\,dx = -\frac{1}{4}\int\frac{dt}{\sqrt{3^2-t^2}} \)
By the standard integral \(\int\frac{dt}{\sqrt{a^2-t^2}} = \sin^{-1}\frac{t}{a}\) with a = 3:
\( \displaystyle = -\frac{1}{4}\sin^{-1}\left(\frac{t}{3}\right)+C \)
\( \displaystyle = -\frac{1}{4}\sin^{-1}\left(\frac{1}{3}\cos^2 2x\right)+C \)
Why the other options are incorrect
Option (1) \(\frac{1}{4}\sin^{-1}\left(\frac{1}{3}\cos^2 2x\right)+C\) — Incorrect. The derivative of \(\cos 2x\) is \(-2\sin 2x\), so the result carries a minus sign.
Option (2) \(\frac{1}{2}\sin^{-1}\left(\frac{1}{3}\cos^2 2x\right)+C\) — Incorrect. Both the sign and the factor are wrong; dt brings \(-4\), giving \(-\frac{1}{4}\).
Option (3) \(-\frac{1}{2}\sin^{-1}\left(\frac{1}{3}\cos^2 2x\right)+C\) — Incorrect. It takes only one factor 2 from the chain rule; there are two (from the square and from 2x).