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Please solve RD Sharma class 12 chapter Indefinite Integrals exercise 18.8 question 45 maths textbook solution

Answers (1)

Answer:

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

Hint:

        cos(A+B)=cosAcosB-sinAsinB

Given:

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

Explanation:

        =\int \! \frac{cos(x+a-b+b)}{sin(x+b)}dx        ...adding and subtracting b in numerator

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

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

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

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

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

Posted by

Gurleen Kaur

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