Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • School
  • Need solution for RD Sharma maths class 12 chapter Indefinite Integrals exercise 18.8 question 47

Need solution for RD Sharma maths class 12 chapter Indefinite Integrals exercise 18.8 question 47

Answers (1)

Answer:

        -\frac{1}{5}log\left | cos5x \right |+\frac{1}{2}log\left | cos2x \right |+\frac{1}{3}log\left | cos3x \right |+C

Hint:

        tan(A+B)=\frac{tanA+tanB}{1-tanA\, tanB}

Given:

        \int \! tan2x\, tan3x\, tan5xdx            .....(1)

Explanation:

        tan5x=tan(2x+3x)=\frac{tan2x+tan3x}{1-tan2x\, tan3x}

Cross-multiplying

        tan5x-tan5x\, tan2x\, tan3x=tan2x+tan3x

        tan5x-tan2x-tan3x=tan5x\, tan2x\, tan3x

Put in (1) we get

        \int \! (tan5x-tan2x-tan3x)dx

        =-\frac{1}{5}log\left | cos5x \right |+\frac{1}{2}log\left | cos2x \right |+\frac{1}{3}log\left | cos3x \right |+C

Posted by

Gurleen Kaur

View full answer

Crack CUET with india's "Best Teachers"

  • HD Video Lectures
  • Unlimited Mock Tests
  • Faculty Support
cuet_ads

Similar Questions

Latest Question