Question

\(\int_{0}^{\pi/4} \sqrt{1 + \sin 2x}\,dx =\)

• A. $1$
• B. $\sqrt{2} + 1$
• C. $\sqrt{2} - 1$
• D. $1 - \sqrt{2}$
• E. $-\sqrt{2}$

Hint:

Whenever you see $1 + \sin 2x$: - Convert it into $(\sin x + \cos x)^2$ - Then simplify the square root easily

Answer 1

The Correct Option is A

Concept: Use trigonometric identity: \[ 1 + \sin 2x = (\sin x + \cos x)^2 \]

Step 1: Simplify the expression inside the square root.
\[ \sqrt{1 + \sin 2x} = \sqrt{(\sin x + \cos x)^2} \] \[ = |\sin x + \cos x| \]

Step 2: Remove modulus using interval.
On $[0, \frac{\pi}{4}]$, both $\sin x$ and $\cos x$ are positive: \[ |\sin x + \cos x| = \sin x + \cos x \]

Step 3: Rewrite the integral.
\[ \int_{0}^{\pi/4} (\sin x + \cos x)\,dx \]

Step 4: Integrate.
\[ = \int_0^{\pi/4} \sin x\,dx + \int_0^{\pi/4} \cos x\,dx \] \[ = \left[-\cos x\right]_0^{\pi/4} + \left[\sin x\right]_0^{\pi/4} \] \[ = \left(-\frac{\sqrt{2}}{2} + 1\right) + \left(\frac{\sqrt{2}}{2} - 0\right) \]

Step 5: Simplify.
\[ = 1 \]