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Need solution for RD Sharma maths class 12 chapter Indefinite Integrals exercise 18.8 question 7

Answers (1)

Answer:

        xcos(b-a)+sin(b-a)log\left | sin(x-b) \right |+C

Hint:

        sin(A+B)=sinA\: cosB+cosA\: sinB

Given:

        \int \! \frac{sin(x-a)}{sin(x-b)}dx

Explanation:

\int \! \frac{sin(x-a)}{sin(x-b)}dx

=\int\frac{sin(x-a+b-b)}{sin(x-b)}dx            [add\; and\; subtract \: b\: in\; (x-a)]

=\int \! \frac{sin(x-b+b-a)}{sin(x-b)}dx

=\int \! \frac{sin(x-b)cos(b-a)+cos(x-b)sin(b-a)}{sin(x-b))}dx     [sin(A+B)=sinA\: cosB+cosA\: sinB]

=\int \frac{\sin (x-b) \cos (b-a)}{\sin (x-b)} d x+\int \frac{\cos (x-b) \sin (b-a)}{\sin (x-b)} d x

=\int\! cos(b-a)dx+\int \! cot(x-b)sin(b-a)dx

=cos(b-a)\int dx+sin(b-a)\int cot(x-b)dx

= cos(b-a)x+ sin(b-a)\: log\left | sin(x-b) \right |+C

[\int\! cot\: x\: dx= log\left | sin\; x \right |+C]

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