#### Explain Solution R.D.Sharma Class 12 Chapter 19 Definite Integrals  Exercise 19.1 Question 62 Maths Textbook Solution.

Answer: $-\sqrt{2}$

Hint: Use indefinite formula then put the limit to solve this integral

Given: $\int_{\pi }^{\frac{3\pi }{2}}\sqrt{1-\cos 2xdx}$

Solution: $\int_{\pi}^{\frac{3 \pi}{2}} \sqrt{1-\cos 2 x} d x=\int_{\pi}^{\frac{3 \pi}{2}} \sqrt{2 \sin ^{2} x} d x \; \; \; \; \; \; \; \: \: \: \: \quad\left[1-\cos 2 \theta=2 \sin ^{2} \theta\right]$

\begin{aligned} &=\int_{\pi}^{\frac{3 \pi}{2}} \sqrt{2} \sin x d x=\sqrt{2} \int_{\pi}^{\frac{3 \pi}{2}} \sin x d x \; \; \; \; \; \; \; \; \; \; \; \; \; \quad\left[\int \sin x d x=-\cos x\right] \\ &=\sqrt{2}[-\cos x]_{\pi}^{\frac{3 \pi}{2}} \\ &=-\sqrt{2}\left[\cos \frac{3 \pi}{2}-\cos \pi\right] \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \quad\left[\cos \left(\pi+\frac{\pi}{2}\right)=-\cos \frac{\pi}{2}\right] \\ &=-\sqrt{2}\left[\cos \left(\pi+\frac{\pi}{2}\right)-\cos \pi\right] \\ &=-\sqrt{2}\left[-\cos \frac{\pi}{2}-\cos \pi\right] \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \quad\left[\begin{array}{c} \cos \frac{\pi}{2}=0 \\ \cos \pi=-1 \end{array}\right] \\ &=-\sqrt{2}[-0-1) \\ &=-\sqrt{2} \times 1 \\ &=-\sqrt{2} \end{aligned}