Evaluate the integral # I = int_(pi/4)^(pi/2) \ 1/sinx \ dx #?

Question

Charles Arocha · Accepted Answer

# int_(pi/4)^(pi/2) \ 1/sinx \ dx = ln (sqrt(2) +1) = 0.88137 # (5dp) 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)

Cameron Alexander · Answer

undefined To evaluate the integral $ I = \int_{\frac{\pi}{4}}^{\frac{\pi}{2}} \frac{1}{\sin(x)} \, dx $, you can use a substitution. Let $ u = \sin(x) $, then $ du = \cos(x) \, dx $.

So the integral becomes $ I = \int_{\frac{\pi}{4}}^{\frac{\pi}{2}} \frac{1}{u} \, du $.

Integrating $ \frac{1}{u} $ with respect to $ u $ yields $ \ln|u| $, so:

$$ I = \ln|\sin(x)| \Bigg|_{\frac{\pi}{4}}^{\frac{\pi}{2}} = \ln|\sin\left(\frac{\pi}{2}\right)| - \ln|\sin\left(\frac{\pi}{4}\right)| $$

$$ I = \ln|1| - \ln\left(\frac{\sqrt{2}}{2}\right) = 0 - \ln\left(\frac{\sqrt{2}}{2}\right) = -\ln\left(\frac{\sqrt{2}}{2}\right) $$