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  • Please solve RD Sharma class 12 chapter Indefinite Integrals exercise 18.8 question 9 maths textbook solution

Please solve RD Sharma class 12 chapter Indefinite Integrals exercise 18.8 question 9 maths textbook solution

Answers (1)

Answer:

        -log\left | cos\: x-sin\: x \right |+C

Hint:

        tan\: x=\frac{sin\: x}{cos\: x}

Given:

        \int \! \frac{1+tan\: x}{1-tan\: x}dx

Explanation:

        \int \! \frac{1+\frac{sin\: x}{cos\: x}}{1-\frac{sin\: x}{cos\: x}}dx=\int \! \frac{cos\: x+sin\: x}{cos\: x-sin\: x}dx

Let,

        cos\: x-sin\: x=t

        (-sin\: x-cos\: x)dx=dt

        =\int \! \frac{dt}{t}

        =-log\left | t \right |

        =-log\left | cos\: x-sin\: x \right |+C

Posted by

Gurleen Kaur

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