JEE Main Mathematics Question (2026) — Solution

Step-by-step solution

Let I = _ - /4 ^ /4 32 ^4 x 1 + e^ x dx Using the definite integral property _ -a ^ a f(x) dx = _ 0 ^ a (f(x) + f(-x)) dx, we get: I = _ 0 ^ /4 ( 32 ^4 x 1 + e^ x + 32 ^4(-x) 1 + e^ (-x) ) dx I = _ 0 ^ /4 ( 32 ^4 x 1 + e^ x + 32 ^4 x 1 + e^ - x ) dx I = _ 0 ^ /4 ( 32 ^4 x 1 + e^ x + 32 ^4 x e^ x e^ x + 1 ) dx I = _ 0 ^ /4 32 ^4 x ( 1 + e^ x 1 + e^ x ) dx I = _ 0 ^ /4 32 ^4 x dx Using the trigonometric identity ^2 x = 1 + 2x 2 , we can write: ^4 x = ( 1 + 2x 2 )^2 = 1 4 (1 + 2 2x + ^2 2x) ^4 x = 1 4 ( 1 + 2 2x + 1 + 4x 2 ) = 3 8 + 1 2 2x + 1 8 4x Substituting this back into the integral: I = 32 _ 0 ^ /4 ( 3 8 + 1 2 2x + 1 8 4x ) dx I = _ 0 ^ /4 (12 + 16 2x + 4 4x) dx Integrating term by term: I = [ 12x + 8 2x + 4x ]_ 0 ^ /4 I = ( 12 ( 4 ) + 8 ( 2 ) + ( ) ) - (0 + 0 + 0) I = 3 + 8(1) + 0 = 3 + 8 Answer: 3 + 8