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  • Need solution for RD Sharma maths Class 12 Chapter 19 Definite Integrals Exercise 19.4 (a) Question 11 textbook solution.

Need solution for  RD Sharma maths Class 12 Chapter 19 Definite Integrals Exercise 19.4 (a) Question 11 textbook solution.

Answers (1)

Answer : \frac{-\pi}{2} \log 2

Given : \int_{0}^{\frac{\pi}{2}}(2 \log \cos x-\log \sin 2 x) d x

Hint : Do integration by parts and you must know properties of log

Solution : I=\int_{0}^{\frac{\pi}{2}}(2 \log \cos x-\log \sin 2 x) d x

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

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

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