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Please solve RD Sharma class 12 chapter Indefinite Integrals exercise 18.8 question 1 maths textbook solution

Answers (1)

Answer:

            \frac{1}{\sqrt{2}}log\left | tan\frac{x}{2} \right |+C

Hint:

            cos2x=1-2sin^{2}x

Given:

           \int \! \frac{1}{\sqrt{1-cos2x}}dx

Explanation:

           \int \! \frac{1}{\sqrt{2sin^{2}\: x}}dx=\frac{1}{\sqrt{2}}\int \frac{1}{sin\: x}dx

        =\frac{1}{\sqrt{2}}\int cos \: ecxdx                                \left [ \int cos \: ecx = log\left | tan\frac{x}{2} \right | \right ]

        =\frac{1}{\sqrt{2}} log\left | tan\frac{x}{2} \right |+C

Posted by

Gurleen Kaur

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