The value of \( \int_{\frac{\pi}{6}}^{\frac{\pi}{3}}\displaystyle\frac{\sin x+\cos x}{\sqrt{\sin 2x}}dx \) is:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \( 2\sin^{-1}\left(\displaystyle\frac{\sqrt{3}-1}{2}\right) \)
The numerator is the derivative of \( \sin x-\cos x \), so the substitution \( t=\sin x-\cos x \) works:
\( \displaystyle dt=(\cos x+\sin x)\,dx \)
\( \displaystyle t^2=1-2\sin x\cos x=1-\sin 2x \)
\( \displaystyle \sin 2x=1-t^2 \)
The limits change to
\( \displaystyle x=\frac{\pi}{6}:\ t=\frac{1}{2}-\frac{\sqrt{3}}{2}=-\frac{\sqrt{3}-1}{2} \)
\( \displaystyle x=\frac{\pi}{3}:\ t=\frac{\sqrt{3}}{2}-\frac{1}{2}=\frac{\sqrt{3}-1}{2} \)
So
\( \displaystyle I=\int_{-\frac{\sqrt{3}-1}{2}}^{\frac{\sqrt{3}-1}{2}}\frac{dt}{\sqrt{1-t^2}} \)
\( \displaystyle =\left[\sin^{-1}t\right]_{-\frac{\sqrt{3}-1}{2}}^{\frac{\sqrt{3}-1}{2}} \)
Since \( \sin^{-1}(-u)=-\sin^{-1}u \),
\( \displaystyle I=2\sin^{-1}\left(\frac{\sqrt{3}-1}{2}\right) \)
Why the other options are incorrect
Option (1) \( \sin^{-1}\left(\displaystyle\frac{\sqrt{3}+1}{2}\right) \) — Incorrect. \( \frac{\sqrt{3}+1}{2}\approx 1.37 \) is greater than 1, so \( \sin^{-1} \) of it is not even defined.
Option (2) \( 2\sin^{-1}\left(\displaystyle\frac{\sqrt{3}+1}{2}\right) \) — Incorrect. \( \frac{\sqrt{3}+1}{2}\approx 1.37 \) is greater than 1, so this is not defined either.
Option (4) \( \sin^{-1}\left(\displaystyle\frac{\sqrt{3}-1}{2}\right) \) — Incorrect. This is only the value at the upper limit; the lower limit contributes an equal amount because \( \sin^{-1} \) is odd.