For \(x \in \left(0,\displaystyle\frac{\pi}{2}\right)\), \(\int\displaystyle\frac{\sin x + \cos x}{\sqrt{\sin 2x}}\,dx\) is equal to
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \(\sin^{-1}(\sin x-\cos x) + C\), C is an arbitrary constant
The numerator is the derivative of sin x − cos x, so let
\(t = \sin x - \cos x,\quad dt = (\cos x + \sin x)\,dx\)
Squaring t:
\(t^2 = \sin^2 x + \cos^2 x - 2\sin x\cos x = 1 - \sin 2x\)
\(\sin 2x = 1 - t^2\)
So
\(\displaystyle \int \frac{\sin x + \cos x}{\sqrt{\sin 2x}}\,dx = \int \frac{dt}{\sqrt{1 - t^2}}\)
\(= \sin^{-1} t + C = \sin^{-1}(\sin x - \cos x) + C\)
Why the other options are incorrect
Option (1) \(\sin^{-1}(\sin 2x)+C\), C is an arbitrary constant — Incorrect. Its derivative is \(\frac{2\cos 2x}{\sqrt{1 - \sin^2 2x}} = \pm 2\), a constant, not the given integrand.
Option (2) \(\sin^{-1}(\cos x-\sin x) + C\), C is an arbitrary constant — Incorrect. Its derivative is \(\frac{-(\sin x + \cos x)}{\sqrt{\sin 2x}}\), the negative of the integrand; the substitution must be t = sin x − cos x, not cos x − sin x.
Option (4) \(\sin^{-1}(\sin x+\cos x) + C\), C is an arbitrary constant — Incorrect. On \(\left(0, \frac{\pi}{2}\right)\), \(\sin x + \cos x = \sqrt{2}\sin\left(x + \frac{\pi}{4}\right) \gt 1\), so \(\sin^{-1}\) of it is not even defined.