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Explain solution RD Sharma class 12 chapter Indefinite Integrals exercise 18.8 question 4 maths

Answers (1)

Answer:

        -2log\left | cos\frac{x}{2} \right |+C

Hint:

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

Given:

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

Explanation:

       \int \sqrt{\frac{2sin^{2}\frac{x}{2}}{2cos^{2}\frac{x}{2}}}dx                    [cos\: x=2cos^{2}\: \frac{x}{2}-1\: \: or\; \; 1-2sin^{2}\frac{x}{2}]

    \int tan\frac{x}{2}dx

Let \frac{x}{2}=t

dx=2dt

2\int tan\: tdt=-2log\left | cos\: t \right |+C

                        =-2log\left | cos\: \frac{x}{2} \right |+C

Posted by

Gurleen Kaur

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