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Question

If $$\displaystyle \int \frac{\cos\: x}{\cos(x-\alpha )}dx=Ax+B\, \log\, \left | \cos(x-\alpha ) \right |+c,$$ then 


A
A=cosα
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B
A=sinα
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C
B=sinα
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D
B=cosα
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Solution

The correct options are
A $$A=\cos \alpha $$
C $$B=\sin\, \alpha $$
$$\displaystyle \int   \dfrac { \cos { x }  }{ \cos  ({ x }-\alpha ) } dx$$
Putting $${ x }-\alpha =\theta $$

$$=\displaystyle \int   \dfrac { \cos  (\alpha +\theta ) }{ \cos  \theta  } { d }\theta $$

$$=\displaystyle \int   \dfrac { \cos  \alpha \cos  \theta -\sin  \alpha \sin  \theta  }{ \cos  \theta  } { d }\theta $$
$$=\cos  \alpha \displaystyle \int  { d } \theta -\sin  \alpha \displaystyle \int   \dfrac { \sin\theta  }{ \cos  \theta  } { d }\theta $$

$$=(\cos  \alpha )\theta +\sin  \alpha \log  |\cos  \theta |+{ K }$$,where$${ K }$$is an arbitrary constant

$$=(\cos  \alpha )({ x }-\alpha )+\sin\alpha \log  |\cos  \theta |+{ K }$$
$$=(\cos  \alpha ){ x }+\sin  \alpha \log  |\cos  ({ x }-\alpha )|+{ C }$$,where$${ C }={ K }-\alpha \cos  \alpha $$is an arbltrary constant.

$$\displaystyle \displaystyle \int  \frac{\cos\: x}{\cos(c-\alpha )}dx=Ax+B\, \log\, \left | cos(x-\alpha ) \right |+c,$$
On comparing, we get

$$\Rightarrow { A }=\cos  \alpha ,{ B }=\sin  \alpha .$$

Mathematics

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