Question

Find the integral of the function~​

Answer 1

Answer:

$\large \bold\red{\frac{1}{ \sin(a - b) } ( ln | \frac{ \cos(x - a) }{ \cos(x - b) } | )+ C}$

Step-by-step explanation:

Given,

$\int \frac{1}{ \cos(x - a) \cos(x - b) } dx$

Now,

Multiply and divide by sin(a-b),

We get,

$= \frac{1}{ \sin(a - b) } \int \frac{ \sin(a - b) }{ \cos(x - a) \cos(x - b) } dx \\ \\ = \frac{1}{ \sin(a - b) } \int \frac{ \sin(a - b + x - x) }{ \cos(x - a) \cos(x - b) } dx \\ \\ = \frac{1}{ \sin(a - b) } \int \frac{ \sin((x - b) - (x - a)) }{ \cos(x - a) \cos(x - b) } dx$

Now,

We know that,

• $\sin( \alpha - \beta ) = \sin( \alpha ) \cos( \beta ) - \cos( \alpha ) \sin( \beta )$

Therefore,

We get,

$= \frac{1}{ \sin(a - b) } \int \frac{ \sin(x - b) \cos(x - a) - \cos(x - b) \sin(x - a) }{ \cos(x - a) \cos(x - b) }dx \\ \\ = \frac{1}{ \sin(a - b) } (\int \frac{ \sin(x - b) }{ \cos(x - b) } - \int \frac{ \sin(x - a) }{ \cos(x - a) } ) dx\\ \\ = \frac{1}{ \sin(a - b) } (\int \tan(x - b) - \int\tan(x - a) ) dx\\ \\ = \frac{1}{ \sin(a - b) } ( - ln | \cos(x - b) | + ln | \cos(x - a) | ) + c \\ \\ = \large \bold{\frac{1}{ \sin(a - b) } ( ln | \frac{ \cos(x - a) }{ \cos(x - b) } | ) + c}$

Answer 2

Answer:

integral of the function

Step-by-step explanation:

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