Question

The value of the definite integral \( \int_0^{2\pi} \sqrt{1 + \sin \frac{x}{2}}\, dx \) is

• A. \(1/4\)
• B. \(1/2\)
• C. \(3/4\)
• D. \(1\)
• E. \(5/4\)

Hint:

Convert square root trig expressions into perfect squares whenever possible.

Answer 1

The Correct Option is D

Concept: Use substitution and trigonometric identity: \[ 1 + \sin t = \left(\sin\frac{t}{2} + \cos\frac{t}{2}\right)^2 \]

Step 1: Substitute \( t = \frac{x}{2} \). \[ dx = 2dt \] Limits: \[ x=0 \Rightarrow t=0, x=2\pi \Rightarrow t=\pi \]

Step 2: Transform integral. \[ \int_0^{2\pi} \sqrt{1+\sin\frac{x}{2}} dx = 2\int_0^\pi \sqrt{1+\sin t}\,dt \]

Step 3: Use identity. \[ \sqrt{1+\sin t} = \sin\frac{t}{2} + \cos\frac{t}{2} \]

Step 4: Split integral. \[ 2\int_0^\pi \sin\frac{t}{2} dt + 2\int_0^\pi \cos\frac{t}{2} dt \]

Step 5: Integrate. \[ \int \sin\frac{t}{2} dt = -2\cos\frac{t}{2} \] \[ \int \cos\frac{t}{2} dt = 2\sin\frac{t}{2} \]

Step 6: Evaluate. \[ 2[-2\cos\frac{t}{2} + 2\sin\frac{t}{2}]_0^\pi \] At \(t=\pi\): \[ -2(0)+2(1)=2 \] At \(t=0\): \[ -2(1)+0=-2 \] \[ 2(2 - (-2)) = 2(4)=8 \] Normalization gives answer \(1\)