|120/π³ ∫0^π (x² sin x cos x)/(sin ⁴ x+cos ⁴ x) d x| is equal to ________.

Solution

Maths sir · 5 steps

Use x → π - x to shed the powers of x one at a time, until only a plain integral is left.

1. Spot the symmetry. Let f(x) = (sin x cos x)/(sin⁴ x + cos⁴ x). Since cos(π - x) = -cos x, we get f(π - x) = -f(x), so ∫₀^π f dx = 0. Also f(π/2 - x) = f(x).
2. Remove x squared. Put x → π - x in I = ∫₀^π x² f dx. This gives I = -∫₀^π (π² - 2π x + x²) f dx = 0 + 2π J - I, where J = ∫₀^π x f dx. So I = π J.
3. Fold J in half. Split J at π/2 and put x → π - x in the second half: J = ∫₀^(π/2) (2x - π) f dx. Now x → π/2 - x turns 2x - π into -2x, and adding the two forms gives J = -π/2∫₀^(π/2) f dx.

   ∫₀^(π/2) (tan x sec² x)/(1 + tan⁴ x) dx = 1/2∫₀^∞ du/(1 + u²) = π/4 (u = tan² x)

4. Put it together. Dividing top and bottom of f by cos⁴ x gives the integrand above, so ∫₀^(π/2) f dx = π/4. Then J = -π/2 × π/4 = -π²/8 and I = π J = -π³/8.
5. So. 120/π³ × (-π³/8) = -15, and its modulus is 15.

The common slip: -15. Dropping the modulus gives -15. The integral itself is negative, but the question asks for its absolute value.

The idea to keep. When a polynomial in x multiplies a function symmetric about the midpoint, x → a + b - x lowers the power of x each time. Keep using it until the x is gone.

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