The value of the integral ∫₀^(π/2) 60 (sin (6 x))/(sin x) d x is equal to _________.

Solution

Maths sir · 4 steps

Break sin 6x into a telescoping sum so the ratio becomes a plain sum of cosines.

1. Telescope the sine. Use sin A - sin B = 2 cos(A+B)/2 sin(A-B)/2. This gives sin 6x - sin 4x = 2 cos 5x sin x, sin 4x - sin 2x = 2 cos 3x sin x and sin 2x = 2 cos x sin x. Add all three and the middle terms cancel.

   (sin 6x)/(sin x) = 2(cos 5x + cos 3x + cos x)

2. Integrate each cosine. ∫₀^(π/2) cos kx dx = 1/k sin kπ/2. So the three terms give 1/5 sin 5π/2 = 1/5, 1/3 sin 3π/2 = -1/3 and sin π/2 = 1.
3. Add up. 1/5 - 1/3 + 1 = (3 - 5 + 15)/15 = 13/15. Multiplying by 2 gives 26/15.
4. So. The integral is 60 × 26/15 = 4 × 26 = 104.

The common slip: 184. Taking sin 3π/2 as +1 instead of -1 makes the sum 23/15 and the answer 184, not 104.

The idea to keep. Whenever you see (sin nx)/(sin x), write sin nx as a telescoping sum. It turns into a sum of cosines, and those integrate in one line each.

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