If _ - 2 ^ 2 dx (1+2024^x)(4 - ^2 2x) = 4 , where is a real constant less than 4 , then the value of is

Step-by-step solution

I = _ - 2 ^ 2 dx (1+2024^x)(4 - ^2 2x) Using the property _ -a ^ a f(x) dx = ₀^ a (f(x) + f(-x)) dx , we get: I = ₀^ 2 ( 1 1+2024^x + 1 1+2024^ -x ) dx 4 - ^2 2x Since 1 1+2024^x + 2024^x 2024^x+1 = 1 , I = ₀^ 2 dx 4 - ^2 2x Let 2x = u dx = du 2 . I = 1 2 ₀^ du 4 - ^2 u Using ₀^ 2a f(x) dx = 2 ₀^ a f(x) dx if f(2a-x) = f(x) , I = ₀^ 2 du 4 - ^2 u Dividing numerator and denominator by ^2 u : I = ₀^ 2 ^2 u , du 4 ^2 u - ^2 u = ₀^ 2 ^2 u , du 4 + (4- ) ^2 u Let u = t ^2 u , du = dt . I = ₀^ dt 4 + (4- )t^2 = 1 4- ₀^ d