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Explain solution RD Sharma class 12 Chapter 19 Definite Integrals Exercise 19.3 question 28

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Answer: \pi^{2}

Hint: You must know about the rules of solving definite integral.

Given:

\int_{0}^{2 \pi} \cos ^{-1}(\cos x) d x

Solution: 

                \cos ^{-1}(\cos x)=\left\{\begin{array}{c} x, 0 \leq x \leq \pi \\ 2 \pi-x, \pi \leq x \leq 2 \pi \end{array}\right\}

                \begin{aligned} &=\int_{0}^{\pi} x d x+\int_{\pi}^{2 \pi}(2 \pi-x) d x \\ & \end{aligned}

                =\left(\frac{x^{2}}{2}\right)_{0}^{\pi}+\left(2 \pi x-\frac{x^{2}}{2}\right)_{\pi}^{2 \pi}

                \begin{aligned} &=\frac{\pi^{2}}{2}+4 \pi^{2}-2 \pi^{2}-2 \pi^{2}+\frac{\pi^{2}}{2} \\ & \end{aligned}

               =\frac{\pi^{2}}{2}+\frac{\pi^{2}}{2}=\pi^{2}

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