Get Answers to all your Questions

header-bg qa

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