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Need solution for RD Sharma maths class 12 chapter Indefinite Integrals exercise 18.8 question 3

Answers (1)

Answer:

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

Hint:

cos\: 2x=2cos^{2}\: x-1=1-2sin^{2}x

Given:

        \int \sqrt{\frac{1+cos\: 2x}{1-cos\: 2x}}dx

Explanation:

    \int \sqrt{\frac{2cos^{2}\: x}{2sin^{2}\: x}}dx                                    \left[\begin{array}{c} 1+\cos 2 x=2 \cos ^{2} x \\ 1-\cos 2x=2 \sin ^{2} x \end{array}\right]

    =\int \sqrt{\left(\frac{\cos x}{\sin x}\right)^{2}} d x

    =\int cot\; xdx

    =log\left | sin\; x \right |+C

Posted by

Gurleen Kaur

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