Compute ∫_0^{π/2} sin^2 x dx.

Compute ∫_0^{π/2} sin^2 x dx.

• A. π/4
• B. π/2
• C. π/8
• D. π

Answer: (A)

The integral can be tackled by using the symmetry between sine and cosine on the interval [0, π/2]. Let I be the integral of sin^2 x over this interval. If we substitute x = π/2 − t, the integral becomes the same as integrating cos^2 t over [0, π/2], so I equals the cosine version as well. Now add the two expressions: 2I = ∫_0^{π/2} (sin^2 x + cos^2 x) dx = ∫_0^{π/2} 1 dx = π/2. Therefore I = π/4. So the value of the integral is π/4.

The integral can be tackled by using the symmetry between sine and cosine on the interval [0, π/2]. Let I be the integral of sin^2 x over this interval. If we substitute x = π/2 − t, the integral becomes the same as integrating cos^2 t over [0, π/2], so I equals the cosine version as well.

Now add the two expressions:

2I = ∫_0^{π/2} (sin^2 x + cos^2 x) dx = ∫_0^{π/2} 1 dx = π/2.

Therefore I = π/4.

So the value of the integral is π/4.