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  • Please solve RD Sharma Class 12 Chapter 19 Definite Integrals Exercise 19.4 (b) Question 27 maths textbook solution.

Please solve RD Sharma Class 12 Chapter 19 Definite Integrals Exercise 19.4 (b) Question 27 maths textbook solution.

Answers (1)

Answer:-  -\pi \log 2

Hints:-  You must know the integral rules of logarithmic functions.

Given:-  \int_{0}^{\pi} \log (1-\cos x) d x

Solution :

\begin{aligned} &I=\int_{0}^{\pi} \log (1-\cos x) d x \\ &=\int_{0}^{\pi} \log \left(2 \sin ^{2} \frac{x}{2}\right) d x \\ &=\int_{0}^{\pi} \log 2 \cdot d x+2 \int_{0}^{\pi} \log \sin \frac{x}{2} \cdot d x \end{aligned}

\begin{array}{ll} \text { Let } t=\frac{x}{2} &\; \; \; \; \; \; x \rightarrow 0 ; t \rightarrow 0 \\ d t=\frac{1}{2} d x & \; \; \; \; \; \; x \rightarrow \pi ; t \rightarrow \frac{\pi}{2} \end{array}

\begin{aligned} &I=\log 2[x]_{0}^{\pi}+4 \int_{0}^{\frac{\pi}{2}} \log \sin t d t \\ &I=\pi \log 2+4 \times\left(\frac{-\pi}{2} \log 2\right) \\ &I=-\pi \log 2 \end{aligned}

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