Question

Evaluate \[ \int_{-\pi/2}^{\pi/2} \sin^2 x\,dx \]

• A. \(\dfrac{\pi}{4}\)
• B. \(\dfrac{\pi}{3}\)
• C. \(\dfrac{\pi}{2}\)
• D. \(\dfrac{3\pi}{4}\)

Hint:

For even functions over symmetric limits, use identities to simplify quickly.

Answer 1

The Correct Option is C

Step 1: Use the identity for \(\sin^2 x\).
\[ \sin^2 x=\frac{1-\cos 2x}{2} \]
Step 2: Substitute into the integral.
\[ \int_{-\pi/2}^{\pi/2}\sin^2 x\,dx =\frac{1}{2}\int_{-\pi/2}^{\pi/2}(1-\cos 2x)\,dx \]
Step 3: Integrate termwise.
\[ =\frac{1}{2}\left[x-\frac{\sin 2x}{2}\right]_{-\pi/2}^{\pi/2} \]
Step 4: Apply limits.
\[ =\frac{1}{2}\left(\pi-0\right)=\frac{\pi}{2} \]