Find

Using the identities 1 = sin²x + cos²x,

sin 2x = 2 sin x cos x

cos 2x = cos²x – sin²x

Using (a + b)² = a² + 2ab + b²

We know that ∫e^x[f(x) + f’(x)] dx = e^x f(x) + c

Here f(x) = sec x hence f’(x) = sec x tan x

Hence ∫e^x [sec x + sec x tan x] dx = e^x sec x + c

⇒ I = (1/2)e^x sec x + c