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Question: Integrate \(\dfrac{{\cos 2x - \cos 2a}}{{\cos x - \cos a}}\) with respect to x....

Integrate cos⁡2x−cos⁡2acos⁡x−cos⁡a\dfrac{{\cos 2x - \cos 2a}}{{\cos x - \cos a}} with respect to x.

Explanation

Solution

Hint: We will use trigonometric identities in the function to eliminate the denominator, which makes it easier to integrate. Also, some formulas for integration will be used. These are given by-
cos⁡2A=2cos⁡2A−1\cos 2{\text{A}} = 2{\cos ^2}{\text{A}} - 1
∫cosxdx=sinx+c\smallint cosxdx = sinx + {\text{c}}

Complete step-by-step answer:

Using the formula for cos 2A, the given expression can be written as-
=2cos⁡2x−1−(2cos⁡2a−1)cosx−cosa= \dfrac{{2{{\cos }^2}{\text{x}} - 1 - \left( {2{{\cos }^2}{\text{a}} - 1} \right)}}{{cosx - cosa}}
=2(cos⁡2x−cos⁡2a)cosx−cosa= \dfrac{{2\left( {{{\cos }^2}{\text{x}} - {{\cos }^2}{\text{a}}} \right)}}{{cosx - cosa}}
=2(cosx−cosa)(cosx+cosa)cosx−cosa= \dfrac{{2\left( {cosx - cosa} \right)\left( {cosx + cosa} \right)}}{{cosx - cosa}}
=2(cosx+cosa)= 2\left( {cosx + cosa} \right)

This is a simplified expression which can be integrated easily. Now, we can integrate the given expression with respect to x,
=∫2(cos⁡x−cos⁡a)dx= \smallint 2\left( {\cos x - \cos a} \right)dx
=2[∫cos⁡xdx−∫cos⁡adx]= 2\left[ {\smallint \cos xdx - \smallint \cos adx} \right]
=2[sin⁡x−xcos⁡a]+c= 2\left[ {\sin x - x\cos a} \right] + c
=2sin⁡x−2xcos⁡a+c= 2\sin x - 2x\cos a + c
Here, cosa is a constant hence it comes out of the integral sign. But cosx is integrated to sinx using the given formula.

This is the required answer.

Note: One common mistake is that the students integrate the cosa into sina, but this is incorrect as cosa is a constant. Also, since limits are not mentioned, we should add an integration constant (c).