If linear functions f(x) and g(x) satisfy int | vedclass

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(A) We are given the integral $I = \int [(1 - 2x)\cos x + (3 + 2x)\sin x] dx$. Using integration by parts,let $u = (1 - 2x)$ and $dv = \cos x dx$,then

Vedclass · Accepted Answer

$f(x) = 3 - 2x$

Vedclass · Answer

(A) We are given the integral $I = \int [(1 - 2x)\cos x + (3 + 2x)\sin x] dx$. Using integration by parts,let $u = (1 - 2x)$ and $dv = \cos x dx$,then $du = -2 dx$ and $v = \sin x$. $\int (1 - 2x)\cos x dx = (1 - 2x)\sin x - \int \sin x (-2) dx = (1 - 2x)\sin x - 2\cos x$. Now,for the second part,let $u = (3 + 2x)$ and $dv = \sin x dx$,then $du = 2 dx$ and $v = -\cos x$. $\int (3 + 2x)\sin x dx = (3 + 2x)(-\cos x) - \int (-\cos x)(2) dx = -(3 + 2x)\cos x + 2\sin x$. Adding both parts: $I = (1 - 2x)\sin x - 2\cos x - (3 + 2x)\cos x + 2\sin x + C$. $I = \sin x(1 - 2x + 2) + \cos x(-2 - 3 - 2x) + C$. $I = (3 - 2x)\sin x + (-5 - 2x)\cos x + C$. Comparing this with $f(x)\sin x + g(x)\cos x + C$,we get $f(x) = 3 - 2x$ and $g(x) = -5 - 2x$. Thus,option $A$ is correct.

If linear functions $f(x)$ and $g(x)$ satisfy $\int {\left[ {\left( {1 - 2x} \right)\cos x + \left( {3 + 2x} \right)\sin x} \right]} dx = f(x)\sin x + g(x)\cos x + C$ (where $C$ is the constant of integration),then which of the following is true?

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