Question

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

A
A=cosα
B
A=sinα
C
B=sinα
D
B=cosα

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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