JEE Main Mathematics Question (2026) — Solution

Step-by-step solution

Let I = _0^ 20 ( ^4 x + ^4 x) \, dx Using the algebraic identity a^2 + b^2 = (a+b)^2 - 2ab, the integrand can be simplified: ^4 x + ^4 x = ( ^2 x + ^2 x)^2 - 2 ^2 x ^2 x Since ^2 x + ^2 x = 1 and 2 x x = 2x: ^4 x + ^4 x = 1 - 1 2 (2 x x)^2 = 1 - 1 2 ^2 2x Using the half-angle formula ^2 = 1 - 2 2 : ^4 x + ^4 x = 1 - 1 2 ( 1 - 4x 2 ) = 1 - 1 4 + 1 4 4x = 3 4 + 1 4 4x Substituting this back into the integral: I = _0^ 20 ( 3 4 + 1 4 4x ) \, dx I = [ 3 4 x + 1 16 4x ]_0^ 20 Evaluating the limits: I = ( 3 4 (20 ) + 1 16 (80 ) ) - (0 + 0) Since (80 ) = 0: I = 15 Answer: 15