Dear Student, ∫π3π21-cos2xdx= ∫π3π22sin2xdx= ∫π3π22sinxdxnow in between π to 3π2 sinx is negative so = -2∫π3π2sinx dx = 2cosxπ3π2= 2cos3π2-cosπ= 20-1-1=2Regards

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D e a r S t u d e n t, \int_{π}^{\frac{3 π}{2}} \sqrt{1 - \cos 2 x} d x = \int_{π}^{\frac{3 π}{2}} \sqrt{2 \sin^{2} x} d x = \int_{π}^{\frac{3 π}{2}} \sqrt{2} |\sin x| d x n o w i n b e t w e e n π to \frac{3 π}{2} sinx is negative so = - \sqrt{2} \int_{π}^{\frac{3 π}{2}} sinx dx = \sqrt{2} {[cosx]}_{π}^{\frac{3 π}{2}} = \sqrt{2} [\cos \frac{3 π}{2} - cosπ] = \sqrt{2} [0 - 1 (- 1)] = \sqrt{2} Regards

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