Question

Help please!

Answer 1

Answer:

$\displaystyle 2 \sin(x) + 2x \cos( \alpha ) + \rm C$

Step-by-step explanation:

we would like to integrate the following integration:

$\displaystyle \int \frac{ \cos(2x) - \cos(2 \alpha ) }{ \cos(x) - \cos( \alpha ) } dx$

notice that you can simplify the integrand

recall that,

$\displaystyle \cos(2 \theta) = 2 \cos(\theta) ^{2} - 1$

thus substitute:

$\displaystyle \int \frac{ 2\cos^{2} (x)- 1 - \{2\cos ^{2} (\alpha ) - 1 \} }{ \cos(x) - \cos( \alpha ) } dx$

remove parentheses:

$\displaystyle \int \frac{ 2\cos^{2} (x)- 1 - 2\cos ^{2} (\alpha ) + 1}{ \cos(x) - \cos( \alpha ) } dx$

$\displaystyle \int \frac{ 2\cos^{2} (x) - 2\cos ^{2} (\alpha ) }{ \cos(x) - \cos( \alpha ) } dx$

factor out 2:

$\displaystyle \int \frac{ 2(\cos^{2} (x) - \cos ^{2} (\alpha )) }{ \cos(x) - \cos( \alpha ) } dx$

we can use algebraic identity i.e

a²-b²=(a+b)(a-b) to factor the denominator

$\displaystyle \int \frac{ 2(\cos^{} (x) + \cos ^{} (\alpha ))( \cos(x) - \cos( \alpha ) ) }{ \cos(x) - \cos( \alpha ) } dx$

reduce fraction:

$\displaystyle \int \frac{ 2(\cos^{} (x) + \cos ^{} (\alpha ))( \cancel{\cos(x) - \cos( \alpha ) ) }}{ \cancel{\cos(x) - \cos( \alpha ) }} dx$

$\displaystyle \int 2( \cos(x) + \cos( \alpha ) )dx$

distribute:

$\displaystyle \int 2 \cos(x) + 2\cos( \alpha ) dx$

use sum integration formula:

$\rm \displaystyle \int 2 \cos(x) dx+ \int2\cos( \alpha ) dx$

recall integration rules:

$\displaystyle 2 \sin(x) + 2x \cos( \alpha )$

and we of course have to add constant of integration

$\displaystyle 2 \sin(x) + 2x \cos( \alpha ) + \rm C$