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  • Provide solution for RD Sharma maths Class 12 Chapter 19 Definite Integrals Exercise 19.4 (b) Question 14 textbook solution.

Provide solution for  RD Sharma maths Class 12 Chapter 19 Definite Integrals Exercise 19.4 (b) Question 14 textbook solution.

Answers (1)

Answer:  -\frac{\pi^{2}}{2} \log (2)

Hints:-  You must know the integration rules of trigonometric and logarithmic functions.

Given:-  \int_{0}^{\pi} x \log \sin x \cdot d x

Solution : \int_{0}^{\pi} x \log \sin x \cdot d x                                            .....(1)

             \begin{aligned} &I=\int_{0}^{\pi}(\pi-x) \log \sin (\pi-x) \cdot d x\\ &I=\int_{0}^{\pi}(\pi-x) \log (\sin x) \cdot d x \end{aligned}             .....(2)

Adding (1) and (2)

            2 I=\pi \int_{0}^{\pi} \log (\sin x) \cdot d x

           \begin{aligned} &2 I=2 \pi \int_{0}^{\pi / 2} \log (\sin x) \cdot d x \\ &I=\pi\left(\frac{-\pi}{2} \log 2\right) \\ &I=\frac{-\pi^{2}}{2} \log (2) \end{aligned}

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