Get Answers to all your Questions

header-bg qa

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}

Posted by


View full answer

Crack CUET with india's "Best Teachers"

  • HD Video Lectures
  • Unlimited Mock Tests
  • Faculty Support