- \(\displaystyle \int_{\frac {\pi}{3}}^{\frac {\pi}{2}} \frac {1-\cos (x)}{\sin (x)} ~dx = (- \ln (\csc (x) + \cot (x))- \ln (\sin (x)))|_{x = \frac {\pi}{3}}^{x = \frac {\pi}{2}} = - \ln (1 + \cos (x))|_{x = \frac {\pi}{3}}^{x = \frac {\pi}{2}} = \color {blue} {\log(\frac {3}{2})}\)
- Solution: \(\frac {1-\cos (x)}{\sin (x)} = \csc (x) - \cot (x)\)
Solution
1. For \(x \in [\pi/3, \pi/2]\), \(\frac{1-\cos(x)}{\sin(x)} = \tan\left(\frac{x}{2}\right) = \csc(x) - \cot(x)\).
1.1. Half-angle identity: \(\tan(x/2) = \frac{1-\cos(x)}{\sin(x)}\).
1.2. Split the fraction: \(\frac{1}{\sin(x)} - \frac{\cos(x)}{\sin(x)} = \csc(x) - \cot(x)\).
2. \(\int \frac{1-\cos(x)}{\sin(x)} \, dx = -2 \ln|\cos(x/2)| + C = -\ln(1+\cos(x)) + C\).
2.1. Using \(\tan(x/2)\): \[\int \tan\left(\frac{x}{2}\right) \, dx = 2 \int \tan(u) \, du = -2 \ln|\cos(u)| + C = -2 \ln\left|\cos\left(\frac{x}{2}\right)\right| + C.\]
2.2. Using the identity \(\cos^2(x/2) = \frac{1+\cos(x)}{2}\), \(-2\ln(\cos(x/2)) = -\ln(\cos^2(x/2)) = -\ln\left(\frac{1+\cos(x)}{2}\right) = -\ln(1+\cos(x)) + \ln(2)\).
2.3. From the form \(\csc x - \cot x\) of step 1, \(\int (\csc x - \cot x)\, dx = -\ln|\csc x + \cot x| - \ln|\sin x| + C = -\ln|(1+\cos x)| + C\).
3. \[\int_{\pi/3}^{\pi/2} \frac{1-\cos(x)}{\sin(x)} \, dx = \left[ -\ln(1+\cos(x)) \right]_{\pi/3}^{\pi/2} = \boxed{\ln\left(\frac{3}{2}\right)}.\]
3.1. Upper limit \(x = \pi/2\): \(\cos(\pi/2) = 0 \implies -\ln(1+0) = -\ln(1) = 0\).
3.2. Lower limit \(x = \pi/3\): \(\cos(\pi/3) = 1/2 \implies -\ln(1+1/2) = -\ln(3/2)\).
3.3. Subtracting lower from upper limit: \[0 - (-\ln(3/2)) = \ln\left(\frac{3}{2}\right).\]