∫e−5x.Cos12x dx=∫eax.Cos bx dx=∫eaxa2+b2a cosx+b sinx+Cnow 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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$\int e^{- 5 x} . C o s 12 x d x =$

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$\frac{e^{- 5 x}}{169} [5 C o s 12 x - 12 S i n 12 x] + c$

$\frac{e^{- 5 x}}{169} [5 C o s 2 x - 12 S i n 12 x] + c$

$- \frac{e^{- 5 x}}{169} [5 C o s 2 x + 12 S i n 12 x] + c$

$- \frac{e^{- 5 x}}{169} [5 C o s 12 x - 12 S i n 12 x] + c$

answer is C.

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Detailed Solution

$\int e^{- 5 x} . C o s 12 x d x$

$= \int e^{a x} . C o s b x d x$

$= \int \frac{e^{a x}}{a^{2} + b^{2}} [a \cos x + b \sin x] + C$

now $a = - 5, b = 12$

$= \frac{e^{- 5 x}}{25 + 144} [- 5 \cos 12 x + 12 \sin 12 x] + C = \frac{e^{- 5 x}}{169} [- 5 \cos 12 x + 12 \sin 12 x] + C = - \frac{e^{- 5 x}}{169} [5 \cos 12 x - 12 \sin 12 x] + C$