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B
(x−a)sina−cosalogsec(x−a)
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C
(x−a)cosa−sinalogsec(x−a)
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D
(x−a)sina+cosalogsec(x−a)
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Solution
The correct option is C(x−a)cosa−sinalogsec(x−a) Let I=∫cosxcos(x−a)dx Put x−a=t⇒x=a+t and dx=dt Therefore I=∫cos(a+t)costdt=∫cosacost−sinasintcostdt =∫(cosa−sinatant)dt=tcosa−sinalogsect =(x−a)cosa−sinalogsec(x−a) Hence, option 'C' is correct.