How do you evaluate the integral #int 1/(1-sinx)dx# from 0 to #pi/2#?

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Answer 1 · 2018-07-04T11:54:18

The integral is divergent.

This is an inproper integral.

Calculate the indefinite integral first by substitution

Let #u=tan(x/2)#

#=>#, #du=1/2sec^2(x/2)dx=1/2(1+u^2)dx#

And

#sinx=(2u)/(1+u^2)#

Therefore, the integral is

#I=int(dx)/(1-sinx)#

#=int(2/(1+u^2))*1/(1-(2u)/(1+u^2))*du#

#=int(2du)/(1+u^2-2u)#

#=int(2du)/(u-1)^2#

#=-2/(u-1)#

#=-2/(tan(x/2)-1)+C#

The definite integral is

#int_0^(y)(dx)/(1-sinx)=[-2/(tan(x/2)-1)]_0^(y)#

#=(-2/(tan(y/2)-1))-(-2/(0-1))#

#=-2-2/(tan(y/2)-1)#

And the improper integral

#lim_(y->pi/2)(-2-2/(tan(y/2)-1))=-oo#

The integral is divergent.

Answer 2 · 2018-07-04T12:14:35

# int 1/(1-sin x) dx =(tan x + sec x)+ C#
# int _0^(pi/2) 1/(1-sin x) dx =oo#

# int 1/(1-sin x) dx =int (1+sin x)/ (1- sin ^2 x) dx#

#int (1+sin x)/cos ^2 x dx = int (sec^2x +tan x sec x) dx#

#= (tan x + sec x)+ C#

# int _0^(pi/2) 1/(1-sin x) dx = [ tan x + sec x]_0^(pi/2)#

#(oo+oo)-(0+1)= (oo-1) = oo# [Ans]