Get Answers to all your Questions

header-bg qa

Please solve RD Sharma class 12 chapter Indefinite Integrals exercise 18.9 question 53 maths textbook solution

Answers (1)


Answer: -\cos (\log x)+c

Hint: Use substitution method to solve this integral.

Given:   \int \frac{\sin (\log x)}{x} d x


        \text { Let } I=\int \frac{\sin (\log x)}{x} d x

        \begin{aligned} &\text { Put } \log x=t \Rightarrow \frac{1}{x} d x=d t \Rightarrow d x=x \; d t \\ &\text { Then } I=\int \frac{\sin t}{x} \cdot x d t=\int \sin t \; d t \end{aligned}

                       \begin{array}{ll} =-\cos t+c & \ \because \sin x \; d x=-\cos x+c] \\ =-\cos (\log x)+c & {[\because t=\log x]} \end{array}

Posted by


View full answer

Crack CUET with india's "Best Teachers"

  • HD Video Lectures
  • Unlimited Mock Tests
  • Faculty Support