integral (sinx+secx)/tanx dx?

Question

i did some steps but im not sure if i did it right
pls help

integral sinx+secx(cotx) dx
integral sinx+1/cosx(cosx/sinx) = so cosx cancel each other= integral sinx+cscx

Answer 1

ʃ (sinx + secx) / tanx dx

= ʃsinx/tanx + secx/tanx dx

= ʃcosx + cscx dx

= sinx - ln |cscx+cotx| + c



Please note: ʃcscx dx = -ln|cscx+cotx|. There is a top secret (!) way to derive this integration formula, and that is to multiply ʃcscx dx by (cscx + cotx)/(cscx + cotx) to get ʃ(csc²x + cscxcotx) / (cscx + cotx) dx, which is ʃ-du/u when letting u = cscx + cotx. That blew my mind when I first learned it!

Answer 2

∫ (sinx + secx)/tanx dx =

∫ sinx/(sinx/cosx) + (1/cosx)/(sinx/cosx) dx =

∫ cosx + cscx dx =

sinx + ln|sin(x/2)| - ln|cos(x/2)| + c

Answer 3

integral (sinx+secx)/tanx dx

=integral (sinx tanx.dx) + integral( secx.tanx.dx)

=integral [(sec(x) - cos(x)dx] + integral[ secx.tanx.dx]

=integral sec(x).dx -integral cos(x).dx +integral[ secx.tanx.dx]

=ln|sec(x) + tan(x)| - sin(x) + C1 + secx + C2



=ln|sec(x) + tan(x)| - sin(x) + secx + C <= ANS