The value of \(\int_{0}^{\frac{\pi}{4}}\displaystyle \sqrt{1 + \sin 2x}\,dx\) is
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 1
Since \(\sin^{2}x + \cos^{2}x = 1\) and \(\sin 2x = 2\sin x\cos x\), the expression is a perfect square:
\( 1 + \sin 2x = \sin^{2}x + \cos^{2}x + 2\sin x\cos x \)
\( = (\sin x + \cos x)^{2} \)
On \([0, \frac{\pi}{4}]\) both sin x and cos x are non-negative, so
\( \sqrt{1 + \sin 2x} = \sin x + \cos x \)
Integrating:
\( \displaystyle \int_{0}^{\frac{\pi}{4}}(\sin x + \cos x)\,dx = \Big[-\cos x + \sin x\Big]_{0}^{\frac{\pi}{4}} \)
\( \displaystyle = \left(-\frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}}\right) - (-1 + 0) \)
\( = 1 \)