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

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

Posted by

Gurleen Kaur

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