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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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