Please help me with the integral (1 / (cosxcotx))?

this is an identity you'll get to memorize!! but here's one way to figure it out

1/cosx = secx and 1/cotx =tanx

so the integral sec(x)tan(x) dx can be found using u substitution

u=sec(x)

du i.e the derivative of sec(x) = tan(x)sec(x) dx

integral 1 du = u + constant

u = sec(x)

so the answer is

sec(x) + constant!!

kind of a silly answer huh! gotta love trig identities!

cotx= cosx /sinx

1/ (cosx cotx)= sin x/cos^2x

INT (1/ (cosx cotx) dx = INT (sinx/cos^2x) dx

u=cosx , du =-sin x dx

= - INT (1/u^2) du = (1/u ) = 1/cosx = sec x + C

cot x = cos x / sin x

cos x cot x = cos² x / sin x

INT ( 1/(cos x cot x))dx = INT (sin x / cos² x) dx

sin x dx = d (- cos x)

INT ( 1/(cos x cot x))dx = INT d(- cos x) / cos² x = - INT dy / y² = 1/y + C = 1/cos x + C = sec x + C

(temporary y = cos x)