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Question 110 of 129

Q.∫e−7xcos⁡x dx\displaystyle\int e^{-7x}\cos x\, dx is:

(a) e−7x50[7cos⁡x−sin⁡x]+c\dfrac{e^{-7x}}{50}[7\cos x - \sin x] + c
(b) e−7x50[−7cos⁡x+sin⁡x]+c\dfrac{e^{-7x}}{50}[-7\cos x + \sin x] + c
(c) e−7x50[7cos⁡x+sin⁡x]+c\dfrac{e^{-7x}}{50}[7\cos x + \sin x] + c
(d) e−7x50[−7cos⁡x−sin⁡x]+c\dfrac{e^{-7x}}{50}[-7\cos x - \sin x] + c
Puducherry TnboardTamil Nadu HSC First Year (DGE) Board 2020MCQ· 1mImportance★★★★★
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Apply the standard reduction formula for ∫eaxcos⁡bx dx\int e^{ax}\cos bx\,dx directly with a=−7, b=1a=-7,\,b=1.

The standard result (derived via integration by parts twice, or by treating this as the real part of ∫e(a+ib)xdx\int e^{(a+ib)x}dx) is:

∫eaxcos⁡bx dx=eax(acos⁡bx+bsin⁡bx)a2+b2+c.\int e^{ax}\cos bx\,dx=\dfrac{e^{ax}(a\cos bx+b\sin bx)}{a^2+b^2}+c.

Here a=−7a=-7, b=1b=1, so a2+b2=49+1=50a^2+b^2=49+1=50. Substituting: …

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