If I=∫₀^(π/2) (sin ^(3/2) x)/(sin ^(3/2) x+cos ^(3/2) x) d x, then ∫₀^(2I) (x sin x cos x)/(sin ⁴ x+cos ⁴ x) d x equals :

Solution

Maths sir · 4 steps

Find I with the a+b−x trick, then use it again to get rid of the x in the second integral.

1. Value of I. Replace x by π/2 - x. This swaps sin and cos, so adding the two forms of I gives 2I = ∫₀^(π/2) 1 dx = π/2. So I = π/4 and the upper limit 2I is π/2.
2. Remove the x. Call the second integral J. Replacing x by π/2 - x leaves the fraction (sin x cos x)/(sin⁴ x + cos⁴ x) unchanged and turns x into π/2 - x. Adding the two forms gives 2J = π/2 K, where K = ∫₀^(π/2) (sin x cos x)/(sin⁴ x + cos⁴ x) dx. So J = π/4 K.

   K = ∫₀^(π/2) (tan x sec² x)/(1 + tan⁴ x) dx = 1/2∫₀^∞ dt/(1+t²) = 1/2 × π/2 = π/4

3. How K works. Divide the numerator and the denominator by cos⁴ x, then put t = tan² x, so dt = 2 tan x sec² x dx. The limits 0 to π/2 become 0 to ∞.
4. So. J = π/4 × π/4 = π²/16, which is option (c).

The idea to keep. When x multiplies a function that is unchanged by x → a + b - x, the property gives ∫ₐᵇ x f(x) dx = (a+b)/2∫ₐᵇ f(x) dx.

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