Question

If linear function $f\left( x \right)$ and $g\left( x \right)$ satisfy $\int {\left[ {\left( {3x - 1} \right)\cos x + \left( {1 - 2x} \right)\sin x} \right]dx = f\left( x \right)\cos x + g\left( x \right) + c}$, then
A.$f\left( x \right) = 3\left( {x - 1} \right)$
B.$f\left( x \right) = 3x - 5$
C.$g\left( x \right) = 3\left( {x - 1} \right)$
D.$g\left( x \right) = 3 + x$

Answer 1

Hint : In order to find the correct option, start with solving the equation on the left side of the function using the product rule, $\smallint u{{ }}v{{ }}dx{{ }} = {{ }}u\smallint v{{ }}dx{{ }} - \smallint u'{{ }}\left( {\smallint v{{ }}dx} \right){{ }}dx$. Solve the two operands of the equations separately and substitute at the original equation at last. Solve and get the results.
Formula used:
$\smallint u{{ }}v{{ }}dx{{ }} = {{ }}u\smallint v{{ }}dx{{ }} - \smallint u'{{ }}\left( {\smallint v{{ }}dx} \right){{ }}dx$
$\int {\cos xdx = \sin x + C_1}$
$\int {\sin xdx = - \cos x + C_2}$

Complete step-by-step answer :
We are given with an equation $\int {\left[ {\left( {3x - 1} \right)\cos x + \left( {1 - 2x} \right)\sin x} \right]dx = f\left( x \right)\cos x + g\left( x \right) + c}$., where $f\left( x \right)$ and $g\left( x \right)$ are two linear functions.
We are initiating with solving the left-hand side of the equation that is $\int {\left[ {\left( {3x - 1} \right)\cos x + \left( {1 - 2x} \right)\sin x} \right]dx}$
We can split the integration into operands and we get:
$\Rightarrow \int {\left( {3x - 1} \right)\cos xdx + \int {\left( {1 - 2x} \right)\sin xdx} }$ ……(1)
We would integrate each operand separately, starting with the first operand:
$\int {\left( {3x - 1} \right)\cos xdx}$
We can see it also contains two different functions inside the bracket, which is $\left( {3x - 1} \right){{ and }}\cos x$, so we would use the product rule of differentiation, that is:
$\smallint u{{ }}v{{ }}dx{{ }} = {{ }}u\smallint v{{ }}dx{{ }} - \smallint u'{{ }}\left( {\smallint v{{ }}dx} \right){{ }}dx$
Comparing $\smallint u{{ }}v{{ }}dx{{ }}$with $\int {\left( {3x - 1} \right)\cos xdx}$, we get:
$u = \left( {3x - 1} \right)$
$v = \cos x$
Substituting the values of u and v in the above equation and we get:
$\Rightarrow \smallint \left( {3x - 1} \right)\cos xdx{{ }} = {{ }}\left( {3x - 1} \right)\smallint {{ }}\cos xdx{{ }} - \smallint \dfrac{{d\left( {3x - 1} \right)}}{{dx}}{{ }}\left( {\smallint \cos x{{ }}dx} \right){{ }}dx$
Since, we know that $\int {\cos xdx = \sin x + C_1}$ and $\dfrac{{d\left( {3x} \right)}}{{dx}} = 3$. So, substituting the values in the above equation and we get:
$\Rightarrow \smallint \left( {3x - 1} \right)\cos xdx{{ }} = {{ }}\left( {3x - 1} \right){{sinx + C_1 }} - \smallint 3{{ sinx }}dx$
$\Rightarrow \smallint \left( {3x - 1} \right)\cos xdx{{ }} = {{ }}\left( {3x - 1} \right){{sinx + C_1 }} - 3\smallint {{sinx }}dx$
Since, we know that $\int {\sin xdx = - \cos x + C_2}$, so substituting it:
$\Rightarrow \smallint \left( {3x - 1} \right)\cos xdx{{ }} = {{ }}\left( {3x - 1} \right){{sinx + C_1 }} - 3\left( { - \cos x + C_2} \right)$
Opening the braces:
$\Rightarrow \smallint \left( {3x - 1} \right)\cos xdx{{ }} = {{ 3xsinx - sinx + C_1 + }}3\cos x + C_2$
$\Rightarrow \smallint \left( {3x - 1} \right)\cos xdx{{ }} = {{ 3xsinx - sinx + }}3\cos x + C_1$ …….(2)
Similarly, solving for the second operand, and we get:
$\int {\left( {1 - 2x} \right)\sin xdx}$
Here also we can see it also contains two different functions inside the bracket, which is $\left( {1 - 2x} \right){{ and sin}}x$, so we would use the product rule of differentiation, that is:
$\smallint u{{ }}v{{ }}dx{{ }} = {{ }}u\smallint v{{ }}dx{{ }} - \smallint u'{{ }}\left( {\smallint v{{ }}dx} \right){{ }}dx$
Comparing $\smallint u{{ }}v{{ }}dx{{ }}$with $\int {\left( {1 - 2x} \right)\sin xdx}$, we get:
$u = \left( {1 - 2x} \right)$
$v = \sin x$
Substituting the values of u and v in the above equation and we get:
$\Rightarrow \int {\left( {1 - 2x} \right)\sin xdx} {{ }} = {{ }}\left( {1 - 2x} \right)\smallint \sin xdx{{ }} - \smallint \dfrac{{d\left( {1 - 2x} \right)}}{{dx}}{{ }}\left( {\smallint \sin x{{ }}dx} \right){{ }}dx$
Since, we know that $\int {\sin xdx = - \cos x + C_3}$ and $\dfrac{{d\left( { - 2x} \right)}}{{dx}} = - 2$. So, substituting the values in the above equation and we get:
$\Rightarrow \int {\left( {1 - 2x} \right)\sin xdx} = {{ }}\left( {1 - 2x} \right)\left( { - \cos x} \right){{ + C_3 }} - \smallint \left( { - 2} \right){{ }}\left( { - \cos x} \right){{ }}dx$
$\Rightarrow \int {\left( {1 - 2x} \right)\sin xdx} = {{ }}\left( {1 - 2x} \right)\left( { - \cos x} \right){{ + C_3 + 2}}\smallint {{ }}\left( { - \cos x} \right){{ }}dx$
$\Rightarrow \int {\left( {1 - 2x} \right)\sin xdx} = {{ }}\left( {1 - 2x} \right)\left( { - \cos x} \right){{ + C_3 - 2}}\smallint {{ }}\cos x{{ }}dx$
Since, we know that $\int {\cos xdx = \sin x + C_4}$, so substituting it:
$\Rightarrow \int {\left( {1 - 2x} \right)\sin xdx} = {{ }}\left( {1 - 2x} \right)\left( { - \cos x} \right){{ + C_3 - 2}}\left( {\sin x + C_4} \right)$
Opening the braces:
$\Rightarrow \int {\left( {1 - 2x} \right)\sin xdx} = {{ - cosx + 2xcosx + C_3 - 2sinx - 2C_4}}$
$\Rightarrow \int {\left( {1 - 2x} \right)\sin xdx} = {{ - cosx + 2xcosx - 2sinx + C_2}}$ …….(3)
Substituting the values of equation 2 and equation 3 in equation 1, we get:
$\Rightarrow \int {\left( {3x - 1} \right)\cos xdx + \int {\left( {1 - 2x} \right)\sin xdx} }$
$\Rightarrow {{3xsinx - sinx + }}3\cos x + C_1 + \left( {{{ - cosx + 2xcosx - 2sinx + C_2}}} \right)$
Solving the equation:
$\Rightarrow {{3xsinx - sinx + }}3\cos x + C_1{{ - cosx + 2xcosx - 2sinx + C_2}}$
$\Rightarrow {{3xsinx - 3sinx + 2}}\cos x{{ + 2xcosx + C}}$
Taking $\sin x$ and $\cos x$ common in braces and we get:
$\Rightarrow \left( {{{3x - 3}}} \right){{sinx + }}\left( {{{2 + 2x}}} \right)\cos x{{ + C}}$
$\Rightarrow \left( {{{2 + 2x}}} \right)\cos x{{ + }}\left( {{{3x - 3}}} \right){{sinx}} + {{C}}$
Comparing this equation with the right-hand side of the equation $\int {\left[ {\left( {3x - 1} \right)\cos x + \left( {1 - 2x} \right)\sin x} \right]dx = f\left( x \right)\cos x + g\left( x \right) + c}$, we get:
$\Rightarrow \left( {{{2 + 2x}}} \right)\cos x{{ + }}\left( {{{3x - 3}}} \right){{sinx}} + {{C = }}f\left( x \right)\cos x + g\left( x \right) + c$
This shows
$f\left( x \right) = \left( {2 + 2x} \right) = 2\left( {1 + x} \right)$
$g\left( x \right) = \left( {3x - 3} \right) = 3\left( {x - 1} \right)$
So, the correct answer is “Option C”.

Note : It’s important to remember the formulas of differentiation and integration to solve this type of question.
We have taken $C_1 + C_2 = C_1$ and other also the same because the sum of two constant terms will also be a constant term.