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

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

Hint:

$sin(A+B)=sinAcosB+cosAsinB$

Given:

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

Explanation:

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

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

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

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

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

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