Evaluate the integral # I = int_(pi/4)^(pi/2) \ 1/sinx \ dx #?
1 Answer
Jun 23, 2017
# int_(pi/4)^(pi/2) \ 1/sinx \ dx = ln (sqrt(2) +1) = 0.88137 # (5dp)
Explanation:
We want to evaluate:
# I = int_(pi/4)^(pi/2) \ 1/sinx \ dx #
# \ \ = int_(pi/4)^(pi/2) \ cscx \ dx #
The integral is a standard which can be looked up, to get
# I = [-ln|cscx+cotx|]_(pi//4)^(pi//2) #
# \ \ = [ln|cscx+cotx|]_(pi//2)^(pi//4) #
# \ \ = ln abs(csc(pi/4) +cot(pi/4)) - ln abs(csc(pi/2) +cot(pi/2))#
# \ \ = ln abs(sqrt(2) +1) - ln abs(1 +0)#
# \ \ = ln (sqrt(2) +1)#
# \ \ = 0.88137 # (5dp)
