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Provide solution for RD Sharma maths class 12 chapter Indefinite Integrals exercise 18.8 question 10

Answers (1)

Answer:

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

Hint:

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

Given:

        \int \! \frac{cosx}{cos(x-a)}dx

Explanation:

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

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

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

=\int \! cos\: a\: dx-\int \! tan(x-a)sin\: a\: dx

\therefore xcos\: a-sin\: a\: log\left | sec(x-a) \right |+C

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Gurleen Kaur

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