Question

please explain it to me that how I gonna answer this : integral of secx I know the answer but how?
It will be ln / secx + tanx / + c​

Answer 1

Answer:

ln|sec(x) + tan(x)| + c

Step-by-step explanation:

∫sec(x) dx

secx = 1/cos(x) = cos(x)/cos²(x) = cos(x)/(1-sin²(x))

let u = sinx

we get ∫secx = ∫du/(1-u²) = ln√(1+u)/(1-u)

as (1+u)/(1-u) = (1+sinx)/(1-sinx) = (sec²x + tan²x)

so ln √(1+u)/(1-u) = ln√(sec²x + tan²x) = ln|secx+tanx| +c