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  • Explain solution RD Sharma class 12 chapter Indefinite Integrals exercise 18.8 question 8 maths

Explain solution RD Sharma class 12 chapter Indefinite Integrals exercise 18.8 question 8 maths

Answers (1)

Answer:

        x\: cos\: cos\: 2a-sin\: sin\: 2a \: log\: log\left | sin\: sin(x+a) \right |+C

Hint:

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

Given:

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

Explanation:

        \int \! \frac{sin(x-a+a-a)}{sin(x+a)}dx                        [Add\; and\; subtract\; a\; in\; (x-a) ]

        =\int \! \frac{sin(x+a-2a)}{sin(x+a)}dx

        =\int \! \frac{sin(x+a)cos2a-cos(x+a)sin2a}{sin(x+a)}dx

        =\int \! \frac{sin(x+a)cos2a}{sin(x+a)}dx-\int \! \frac{cos(x+a)sin2a}{sin(x+a)}dx

        =\int \!cos2a\, dx-\int \! cot(x+a)sin2a\, dx

        =cos2a\int \! dx-sin2a\int cot(x+a)dx

        =cos2ax-sin2a\; log\left | sin(x+a) \right |+C

        =xcos2a-sin2a\; log\left | sin(x+a) \right |+C

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Gurleen Kaur

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