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Provide solution for RD Sharma maths class 12 chapter Indefinite Integrals exercise 18.8 question 14

Answers (1)

Answer:

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

Hint:

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

Given:

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

Explanation:

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

 Let

        sin\: x+cos\: x=t

        (cos\: x-sin\: x)dx=dt

        =-\int \! \frac{dt}{t}\; \; \; \; =-log\: t+C

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

Posted by

Gurleen Kaur

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