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  • Provide solution for RD Sharma maths class 12 chapter Indefinite Integrals exercise 18.8 question 2

Provide solution for RD Sharma maths class 12 chapter Indefinite Integrals exercise 18.8 question 2

Answers (1)

Answer:

        \sqrt{2}log\left | tan\left ( \frac{\pi }{4}+\frac{x}{4} \right ) \right |+C

Hint:

        cos 2\: x=2cos^{2}\: x-1

Given:

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

Explanation:

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

    =\frac{1}{\sqrt{2}}\int \frac{1}{cos\frac{x}{2}}dx

    =\frac{1}{\sqrt{2}}\int \! sec\frac{x}{2}dx

    Let \frac{x}{2}=t

    dx=2dt

                        =\frac{2}{\sqrt{2}} \int sec\: t\: dt =\sqrt{2} \int sec\: t\: dt

                        =\sqrt{2}\: log\left | tan \left ( \frac{\pi }{4}+\frac{t}{2} \right ) \right |+C           \because \int\! sec\: x\: dx= log\: \left | tan \left ( \frac{\pi }{4}+\frac{x}{2} \right ) \right |  

                        =\sqrt{2}\: log\! \left | tan \left ( \frac{\pi }{4}+\frac{x}{2} \right ) \right |+C  

Posted by

Gurleen Kaur

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