Question

Let $\int {g\left( x \right)} dx = F\left( x \right)$. Then, $\int {{x^3}g\left( {{x^2}} \right)} dx =$
(a) $\dfrac{1}{2}\left[ {{x^2}{{\left( {F\left( x \right)} \right)}^2} - \int {{{\left( {F\left( x \right)} \right)}^2}} dx} \right]$
(b) $\dfrac{1}{2}\left[ {{x^2}F{{\left( x \right)}^2} - \int {\left( {F\left( {{x^2}} \right)} \right)} d\left( {{x^2}} \right)} \right]$
(c) $\dfrac{1}{2}\left[ {{x^2}F\left( x \right) - \dfrac{1}{2}\int {{{\left( {F\left( x \right)} \right)}^2}} dx} \right]$
(d) None of these

Answer 1

Here, we need to find the value of the given integral. We will use a substitution method to simplify the expression to be integrated. Then, we will use integration by parts and the given information to find the required value of the integral.

Formula Used:
We will use the formula of integration by parts, the integral of the product of two differentiable functions of $x$ can be written as $\int {uv} dx = u\int v dx - \int {\left( {\dfrac{{d\left( u \right)}}{{dx}} \times \int v dx} \right)dx}$, where $u$ and $v$ are the differentiable functions of $x$.

Complete step-by-step answer:
We will use substitution method to simplify the given integral.
Rewriting the given integral, we get
$\Rightarrow \int {{x^3}g\left( {{x^2}} \right)} dx = \int {x \cdot {x^2}g\left( {{x^2}} \right)} dx$
Let ${x^2} = t$.
Differentiate both sides with respect to $x$, we get
$\Rightarrow 2x = \dfrac{{dt}}{{dx}}$
Multiplying both sides by $dx$, we get
$\Rightarrow 2xdx = dt$
Dividing both sides by 2, we get
$\Rightarrow xdx = \dfrac{1}{2}dt$
Substituting ${x^2} = t$ and $xdx = \dfrac{1}{2}dt$ in the equation $\int {{x^3}g\left( {{x^2}} \right)} dx = \int {x \cdot {x^2}g\left( {{x^2}} \right)} dx$, we get
$\Rightarrow \int {{x^3}g\left( {{x^2}} \right)} dx = \int {\dfrac{1}{2}tg\left( t \right)} dt$
Therefore, we get
$\Rightarrow \int {{x^3}g\left( {{x^2}} \right)} dx = \dfrac{1}{2}\int {tg\left( t \right)} dt$
Thus, we have simplified the expression within the integral.
Now, we will integrate the simplified function using integration by parts.
Using integration by parts, the integral of the product of two differentiable functions of $x$ can be written as $\int {uv} dx = u\int v dx - \int {\left( {\dfrac{{d\left( u \right)}}{{dx}} \times \int v dx} \right)dx}$, where $u$ and $v$ are the differentiable functions of $x$.
Let $u$ be $t$ and $v$ be $g\left( t \right)$.
Therefore, by integrating $\int {tg\left( t \right)} dt$ by parts, we get
\Rightarrow \int {tg\left( t \right)} dt = t\int {\left[ {g\left( t \right)} \right]} dt - \int {\left[ {\dfrac{{d\left( t \right)}}{{dt}} \times \int {\left\\{ {g\left( t \right)} \right\\}} dt} \right]} dt
The derivative of a variable with respect to itself is always 1.
Thus, we get
\Rightarrow \int {tg\left( t \right)} dt = t\int {\left[ {g\left( t \right)} \right]} dt - \int {\left[ {1 \times \int {\left\\{ {g\left( t \right)} \right\\}} dt} \right]} dt
Multiplying the terms, we get
\Rightarrow \int {tg\left( t \right)} dt = t\int {\left[ {g\left( t \right)} \right]} dt - \int {\left[ {\int {\left\\{ {g\left( t \right)} \right\\}} dt} \right]} dt
Now, it is given that $\int {g\left( x \right)} dx = F\left( x \right)$.
We know that changing the variable would not change the value of the integral. This can be written as $\int {f\left( x \right)} dx = \int {f\left( t \right)} dt$.
Therefore, substituting $x = t$ in the equation $\int {g\left( x \right)} dx = F\left( x \right)$, we get
$\Rightarrow \int {g\left( t \right)} dt = F\left( t \right)$
Substituting $\int {g\left( t \right)} dt = F\left( t \right)$ in the equation \int {tg\left( t \right)} dt = t\int {\left[ {g\left( t \right)} \right]} dt - \int {\left[ {\int {\left\\{ {g\left( t \right)} \right\\}} dt} \right]} dt, we get
$\Rightarrow \int {tg\left( t \right)} dt = t \times F\left( t \right) - \int {\left[ {F\left( t \right)} \right]} dt$
Substituting $\int {tg\left( t \right)} dt = t \times F\left( t \right) - \int {\left[ {F\left( t \right)} \right]} dt$ in the equation $\int {{x^3}g\left( {{x^2}} \right)} dx = \dfrac{1}{2}\int {tg\left( t \right)} dt$, we get
$\begin{array}{l} \Rightarrow \int {{x^3}g\left( {{x^2}} \right)} dx = \dfrac{1}{2}\left[ {t \times F\left( t \right) - \int {\left( {F\left( t \right)} \right)} dt} \right]\\\ \Rightarrow \int {{x^3}g\left( {{x^2}} \right)} dx = \dfrac{1}{2}\left[ {tF\left( t \right) - \int {\left( {F\left( t \right)} \right)} dt} \right]\end{array}$
Finally, substituting $t = {x^2}$, we get
$\Rightarrow \int {{x^3}g\left( {{x^2}} \right)} dx = \dfrac{1}{2}\left[ {{x^2}F\left( {{x^2}} \right) - \int {\left( {F\left( {{x^2}} \right)} \right)} d\left( {{x^2}} \right)} \right]$
Rewriting the equation, we get
$\Rightarrow \int {{x^3}g\left( {{x^2}} \right)} dx = \dfrac{1}{2}\left[ {{x^2}F{{\left( x \right)}^2} - \int {\left( {F\left( {{x^2}} \right)} \right)} d\left( {{x^2}} \right)} \right]$
Therefore, we get the value of the integral $\int {{x^3}g\left( {{x^2}} \right)} dx$ as $\dfrac{1}{2}\left[ {{x^2}F{{\left( x \right)}^2} - \int {\left( {F\left( {{x^2}} \right)} \right)} d\left( {{x^2}} \right)} \right]$.
Thus, the correct option is option (b).

Note: We differentiated ${x^2}$ as $2x$. This is because the derivative of a function of the form ${x^n}$ is $n{x^{n - 1}}$. Thus, the derivative of the expression ${x^2}$ is $2{x^1} = 2x$.
We simplified $\int {\dfrac{1}{2}tg\left( t \right)} dt$ as $\dfrac{1}{2}\int {tg\left( t \right)} dt$. This is because the integral of a function of the form $af\left( x \right)$ can be written as $\int {af\left( x \right)} dx = a\int {f\left( x \right)} dx$.