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  • Please solve RD Sharma class 12 chapter Indefinite Integrals exercise 18.8 question 49 maths textbook solution

Please solve RD Sharma class 12 chapter Indefinite Integrals exercise 18.8 question 49 maths textbook solution

Answers (1)

Answer:

        log\left | sin(x+\frac{\pi }{6})sin(x-\frac{\pi }{6}) \right |+C

Hint:

        sin(A+B)=sinAcosB+cosAsinB

Given:

        \int \! \frac{sin2x}{sin(x-\frac{\pi }{6})sin(x+\frac{\pi }{6})}dx

Explanation:

        \int \! \frac{sin[(x-\frac{\pi }{6}+\frac{\pi }{6}+x)]}{sin(x-\frac{\pi }{6})sin(x+\frac{\pi }{6})}dx

        \int \! \frac{sin[(x-\frac{\pi }{6})+(x+\frac{\pi }{6})]}{sin(x-\frac{\pi }{6})sin(x+\frac{\pi }{6})}dx

        =\int \frac{\sin \left(x-\frac{\pi}{6}\right) \cos \left(x+\frac{\pi}{6}\right)+\cos \left(x-\frac{\pi}{6}\right) \sin \left(x+\frac{\pi}{6}\right)}{\sin \left(x-\frac{\pi}{6}\right) \sin \left(x+\frac{\pi}{6}\right)}

        =\int\! cot(x+\frac{\pi }{6})dx+\int \! cot(x-\frac{\pi }{6})dx

        =log\left | sin(x+\frac{\pi }{6}) \right |+log\left | sin(x-\frac{\pi }{6}) \right |+C

        =log\left | sin(x+\frac{\pi }{6})sin(x-\frac{\pi }{6}) \right |+C

Posted by

Gurleen Kaur

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