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  • 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

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