The value of
∫₀^(2π) (xsin ⁸x)/(sin ⁸x + cos ⁸x) dx is equal to :

Solution

Maths sir · 4 steps

Flip x to 2π - x so the stray x becomes a constant, then pair sin⁸ with cos⁸.

1. Remove the x. Let f(x) = (sin⁸ x)/(sin⁸ x + cos⁸ x). Since sin(2π - x) = -sin x and cos(2π - x) = cos x, the eighth powers do not change, so f(2π - x) = f(x). Replacing x by 2π - x turns I = ∫₀^(2π) x f(x) dx into ∫₀^(2π) (2π - x) f(x) dx.

   2I = 2π ∫₀^(2π) f(x) dx ⇒ I = π ∫₀^(2π) f(x) dx

2. Use the period. f(x + π) = f(x), and on [0, π] the function is symmetric about π/2. So ∫₀^(2π) f dx = 4∫₀^(π/2) f dx.
3. Pair the integrals. Call J = ∫₀^(π/2) f dx. Replacing x by π/2 - x swaps sin and cos, so J = ∫₀^(π/2) (cos⁸ x)/(sin⁸ x + cos⁸ x) dx. Adding the two forms gives 2J = ∫₀^(π/2) 1 dx = π/2, so J = π/4.
4. So. ∫₀^(2π) f dx = 4 × π/4 = π, and I = π × π = π². The answer is (c).

The idea to keep. If f(a - x) = f(x), then ∫₀ᵃ x f(x) dx = a/2∫₀ᵃ f(x) dx. After that, sinⁿ/(sinⁿ + cosⁿ) over [0, π/2] always gives π/4.

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