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Question

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


A
x cos(ba) sin(ba)log| sec(xb)|+c
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B
x cos(ba)+ sin(ba)log| sec(xb)|+c
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C
x cos(b+a) sin(b+a)log| sec(x+b)|+c
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D
x cos(b+a)+ sin(b+a)log| sec(xb)|+c
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Solution

The correct option is D $$ x \ cos(b-a) - \ sin(b-a) log|\ sec(x-b)| + c$$
$$\dfrac{\cos (x-a)}{\cos (x-b)}=\dfrac{\cos (x-b+(b-a))}{\cos (x-b)}=\cos (b-a)-\tan (x-b)\sin (b-a)$$

$$\displaystyle\int \dfrac{\cos (x-a)}{\cos (x-b)}d{x}=\cos (b-a)\int  d{x}-\sin (b-a)\int \tan (x-b)d{x}=x\cos (b-a)-\sin (b-a)\log |\text{sec}(x-b)|+C$$


Mathematics

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