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  • Provide Solution for RD Sharma Class 12 Chapter 19 Definite Integrals Exercise 19.3 Question 25

Provide Solution for RD Sharma Class 12 Chapter 19 Definite Integrals Exercise 19.3 Question 25

Answers (1)

Answer:   \frac{\pi^{2}}{8}

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

Given: 

                \int_{\frac{-\pi}{2}}^{\pi} \sin ^{-1}(\sin x) d x

Solution:

                \begin{aligned} &I=\int_{\frac{-\pi}{2}}^{\frac{\pi}{2}} \sin ^{-1}(\sin x) d x \\ \end{aligned} `

                \sin ^{-1}(\sin x)=\left\{\begin{array}{l} x, \quad \frac{-\pi}{2} \leq x \leq \frac{\pi}{2} \\\\ (\pi-x), \frac{\pi}{2} \leq x \leq \frac{3 \pi}{2} \end{array}\right\}

                \begin{aligned} &I=\int_{\frac{-\pi}{2}}^{\frac{\pi}{2}} \sin ^{-1}(\sin x) d x+\int_{\frac{\pi}{2}}^{\pi} \sin ^{-1}(\sin x) d \\ & \end{aligned}

                I=\int_{\frac{-\pi}{2}}^{\frac{\pi}{2}} x d x+\int_{\frac{\pi}{2}}^{\pi}(\pi-x) d x                             \quad\left[\sin ^{-1}(\sin x)=x\right]

                \begin{aligned} &I=\left(\frac{x^{2}}{2}\right)_{\frac{-\pi}{2}}^{\frac{\pi}{2}}+\left(\pi x-\frac{x^{2}}{2}\right)_{\frac{\pi}{2}}^{\pi} \\ \end{aligned}

                I=\frac{1}{2}\left(\frac{\pi^{2}}{4}-\frac{\pi^{2}}{4}\right)+\left(\pi \times \pi-\frac{\pi^{2}}{2}-\left(\pi \times \frac{\pi}{2}-\frac{\frac{\pi^{2}}{4}}{2}\right)\right) \\

                =\pi^{2}-\frac{\pi^{2}}{2}-\frac{\pi^{2}}{2}+\frac{\pi^{2}}{8}=\frac{\pi^{2}}{8}

 

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