Question

The value of \[ \frac{4}{\pi} \int_0^{\pi/2} \sin^2 x \, dx \] is ________________ (rounded off to two decimal places).

Hint:

Integrals of $\sin^2 x$ or $\cos^2 x$ over a full or half period often reduce to simple fractions of $\pi$. Use the half-angle identity.

Answer 1

Correct Answer: 1

Step 1: Use trigonometric identity.
\[ \sin^2 x = \frac{1 - \cos 2x}{2} \]
Step 2: Substitute in integral.
\[ I = \int_0^{\pi/2} \sin^2 x \, dx = \int_0^{\pi/2} \frac{1 - \cos 2x}{2} \, dx \] \[ = \frac{1}{2} \int_0^{\pi/2} dx - \frac{1}{2} \int_0^{\pi/2} \cos 2x \, dx \]
Step 3: Evaluate separately.
\[ \int_0^{\pi/2} dx = \frac{\pi}{2} \] \[ \int_0^{\pi/2} \cos 2x \, dx = \left[\frac{\sin 2x}{2}\right]_0^{\pi/2} = \frac{\sin \pi}{2} - \frac{\sin 0}{2} = 0 \]
Step 4: Final value of integral.
\[ I = \frac{1}{2} . \frac{\pi}{2} = \frac{\pi}{4} \]
Step 5: Multiply with coefficient.
\[ \frac{4}{\pi} . I = \frac{4}{\pi} . \frac{\pi}{4} = 1 \] Final Answer: \[ \boxed{1.00} \]