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Question

Evaluate: cosxcos(xa)dx

A
(xa)cosa+sinalogsec(xa)
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B
(xa)sinacosalogsec(xa)
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C
(xa)cosasinalogsec(xa)
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D
(xa)sina+cosalogsec(xa)
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Solution

The correct option is C (xa)cosasinalogsec(xa)
Let I=cosxcos(xa)dx
Put xa=tx=a+t and dx=dt
Therefore
I=cos(a+t)costdt=cosacostsinasintcostdt
=(cosasinatant)dt=tcosasinalogsect
=(xa)cosasinalogsec(xa)
Hence, option 'C' is correct.

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