Q.

e5x.Cos12xdx=

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a

e-5x1695 Cos2x-12Sin12x+c

b

e-5x1695 Cos12x-12Sin12x+c

c

-e-5x1695 Cos12x-12Sin12x+c

d

-e-5x1695 Cos2x+12Sin12x+c

answer is C.

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Detailed Solution

e5x.Cos12xdx

=eax.Cos bxdx

=eaxa2+b2a cosx+b sinx+C

now a=-5, b=12

=e-5x25+144-5 cos12x+12 sin 12x+C =e-5x169-5 cos12x+12sin 12x+C =-e-5x1695 cos12x-12 sin 12x+C

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∫e−5x.Cos12x dx=