Answer
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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-
$\cos 2{\text{A}} = 2{\cos ^2}{\text{A}} - 1$
$\smallint cosxdx = sinx + {\text{c}}$
Complete step-by-step answer:
Using the formula for cos 2A, the given expression can be written as-
$ = \dfrac{{2{{\cos }^2}{\text{x}} - 1 - \left( {2{{\cos }^2}{\text{a}} - 1} \right)}}{{cosx - cosa}}$
$ = \dfrac{{2\left( {{{\cos }^2}{\text{x}} - {{\cos }^2}{\text{a}}} \right)}}{{cosx - cosa}}$
$ = \dfrac{{2\left( {cosx - cosa} \right)\left( {cosx + cosa} \right)}}{{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,
$ = \smallint 2\left( {\cos x - \cos a} \right)dx$
$ = 2\left[ {\smallint \cos xdx - \smallint \cos adx} \right]$
$ = 2\left[ {\sin x - x\cos a} \right] + 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).
$\cos 2{\text{A}} = 2{\cos ^2}{\text{A}} - 1$
$\smallint cosxdx = sinx + {\text{c}}$
Complete step-by-step answer:
Using the formula for cos 2A, the given expression can be written as-
$ = \dfrac{{2{{\cos }^2}{\text{x}} - 1 - \left( {2{{\cos }^2}{\text{a}} - 1} \right)}}{{cosx - cosa}}$
$ = \dfrac{{2\left( {{{\cos }^2}{\text{x}} - {{\cos }^2}{\text{a}}} \right)}}{{cosx - cosa}}$
$ = \dfrac{{2\left( {cosx - cosa} \right)\left( {cosx + cosa} \right)}}{{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,
$ = \smallint 2\left( {\cos x - \cos a} \right)dx$
$ = 2\left[ {\smallint \cos xdx - \smallint \cos adx} \right]$
$ = 2\left[ {\sin x - x\cos a} \right] + 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).
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