Answer:
[tex] \displaystyle 2 \sin(x) + 2x \cos( \alpha ) + \rm C[/tex]
Step-by-step explanation:
we would like to integrate the following integration:
[tex] \displaystyle \int \frac{ \cos(2x) - \cos(2 \alpha ) }{ \cos(x) - \cos( \alpha ) } dx[/tex]
notice that you can simplify the integrand
recall that,
[tex] \displaystyle \cos(2 \theta) = 2 \cos(\theta) ^{2} - 1[/tex]
thus substitute:
[tex]\displaystyle \int \frac{ 2\cos^{2} (x)- 1 - \{2\cos ^{2} (\alpha ) - 1 \} }{ \cos(x) - \cos( \alpha ) } dx[/tex]
remove parentheses:
[tex]\displaystyle \int \frac{ 2\cos^{2} (x)- 1 - 2\cos ^{2} (\alpha ) + 1}{ \cos(x) - \cos( \alpha ) } dx[/tex]
[tex]\displaystyle \int \frac{ 2\cos^{2} (x) - 2\cos ^{2} (\alpha ) }{ \cos(x) - \cos( \alpha ) } dx[/tex]
factor out 2:
[tex]\displaystyle \int \frac{ 2(\cos^{2} (x) - \cos ^{2} (\alpha )) }{ \cos(x) - \cos( \alpha ) } dx[/tex]
we can use algebraic identity i.e
a²-b²=(a+b)(a-b) to factor the denominator
[tex]\displaystyle \int \frac{ 2(\cos^{} (x) + \cos ^{} (\alpha ))( \cos(x) - \cos( \alpha ) ) }{ \cos(x) - \cos( \alpha ) } dx[/tex]
reduce fraction:
[tex]\displaystyle \int \frac{ 2(\cos^{} (x) + \cos ^{} (\alpha ))( \cancel{\cos(x) - \cos( \alpha ) ) }}{ \cancel{\cos(x) - \cos( \alpha ) }} dx[/tex]
[tex] \displaystyle \int 2( \cos(x) + \cos( \alpha ) )dx[/tex]
distribute:
[tex] \displaystyle \int 2 \cos(x) + 2\cos( \alpha ) dx[/tex]
use sum integration formula:
[tex] \rm \displaystyle \int 2 \cos(x) dx+ \int2\cos( \alpha ) dx[/tex]
recall integration rules:
[tex] \displaystyle 2 \sin(x) + 2x \cos( \alpha ) [/tex]
and we of course have to add constant of integration
[tex] \displaystyle 2 \sin(x) + 2x \cos( \alpha ) + \rm C[/tex]
