Question

Hence evaluate:

`int_(-2π)^(2π) (sin^4x + cos^4x)/(1 + e^x)dx`

Answer 1

I = `int_(-2π)^(2π) (sin^4x + cos^4x)/(1 + e^x)dx`

= `int_(-2π)^(2π) (sin^4(-x) + cos^4(-x))/(1 + e^-x)dx`  ...`[∵ int_a^b f(x)dx = int_a^b f(a + b - x)dx]`

= `int_(-2π)^(2π) (sin^4x + cos^4x)/(1 + e^-x)dx`

= `int_(-2π)^(2π) (e^x(sin^4x + cos^4x))/(1 + e^x)dx`

I + I = `int_(-2π)^(2π) (sin^4x + cos^4x)/(1 + e^x)dx + int_(-2π)^(2π) (e^x(sin^4x + cos^4x))/(1 + e^x)dx`

∴ 2I = `int_(-2π)^(2π) (sin^4x + cos^4x)dx`

= `2int_0^(2π) (sin^4x +  cos^4x)dx`  ...`[∵ int_-a^a f(x)dx = 2int_0^a f(x)dx  "if"  f(-x) = f(x)]`

∴ I = `int_0^(2π) (sin^4x + cos^4x)dx`

= `2int_0^π (sin^4x + cos^4x)dx`  ...`[∵ int_0^(2a) f(x)dx = 2int_0^af(x)dx  "if"  f(2a - x) = f(x)]`

= `2*(3π)/4`

= `(3π)/2`