what 's the integral of cos^1/2(x)/sin(x) ?

Answer 1

u= cos^(1/2)x du = 1/2 cos^(-1/2)x * (-sin x) dx

dx = 2 cos^(1/2)x/ (-sin x) du-----------------(1)

integral i = cos^(1/2)x/sinx dx = cos^(1/2)x/sinx * 2cos(1/2)x/(-sinx)du = -2cosx /sin^2(x)du ----------(2)

cosx =u^2 sin^2(x)= 1- cos ^2(x) - 1- u^4

i= -2u^2/(1-u^4) du = (1/(1+u^2) - 1/(1-u^2))du

now each of these integrals 1/1+u^2 and 1/ (1-u^2) could be solved easily 1/1+u^2) could be tackled by u = tant giving
arctan u

and the other one 1/1-u^2) could be taken as 1/(1-u) + 1/ (1+u) giving log(1+u)/(1-u)

now substitute u as cos ^(1/2)x !!!!!

Answer 2

write in the box

sqrt(cos(x))/sin(x)

Then you will get the answer:

tan^-1(sqrt(cosx)+1/2log(sqrt(cosx)-1)-1/2log(sqrt(cosx)+1)