∫₀^(20 π)(|sin x|+|cos x|)² d x is equal to

Solution

Maths sir · 4 steps

Square it first: the bracket becomes 1 + |sin 2x|, and that repeats every π/2.

1. Expand the square. (|sin x| + |cos x|)² = sin² x + cos² x + 2|sin x cos x| = 1 + |sin 2x|.

   ∫₀^(20π)(1 + |sin 2x|)dx = 20π + ∫₀^(20π)|sin 2x| dx

2. Use the period. |sin 2x| has period π/2, so 0 to 20π covers 20π/(π/2) = 40 periods. Over one period, ∫₀^(π/2)sin 2x dx = [-(cos 2x)/2]₀^(π/2) = 1.
3. Add up. The second integral is 40 × 1 = 40, so the total is 20π + 40.
4. So. The integral equals 20π + 40 = 20(π + 2), which is option (d).

The idea to keep. For a periodic integrand over a long range, find the period, integrate over one period, and multiply by the number of periods. Absolute values usually shrink the period.

Answer from the NTA final answer key. Solution by PrepWiser.