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Nuetrik [128]
3 years ago
14

Help please!

Mathematics
1 answer:
Degger [83]3 years ago
5 0

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

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