What is th integral of (1 + sinx) / ((cosx)^2) from 0 to pi/4?

Question

Urgent!

Answer 1

∫(1+sin x)/cos² x dx
∫sec² x dx + ∫sin x/cos² x dx
tan² x + ∫sin x/cos² x dx
u=cos x, du=- sin x dx
tan² x - ∫1/u² du
tan² x + 1/u + C
tan² x + sec x + C

Evaluating from 0 to π/4

tan² (π/4) + sec (π/4) - tan² 0 - sec 0
1 + √2 - 0 - 1
√2

Answer 2

(1 + sinx)/ cos^2x
=1/cos^2x + sinx/cos^2x
=sec^2x +secx.tanx

integrating the whole thing u get
tan x -secx
putting limits ,u get
=(1-root2) - (0-1)
=(2-root2)
just check it once again .....
if i have dun sum mistake, pl forgive
(im a human ,nd doin a mistake is his inherent property)