Question

If $\int {f(x)dx} = \Psi (x)$, then $\int {{x^5}f\left( {{x^3}} \right)dx} $ is equal to:
A. $\dfrac{1}{3}{x^3}\Psi \left( {{x^3}} \right) - 3\int {{x^3}\Psi \left( {{x^3}} \right)dx} + C$
B. $\dfrac{1}{3}{x^3}\Psi \left( {{x^3}} \right) - \int {{x^2}\Psi \left( {{x^3}} \right)dx} + C$
C. $\dfrac{1}{3}\left[ {{x^3}\Psi \left( {{x^3}} \right) - \int {{x^3}\Psi \left( {{x^3}} \right)dx} } \right] + C$
D. $\dfrac{1}{3}\left[ {{x^3}\Psi \left( {{x^3}} \right) - \int {{x^2}\Psi \left( {{x^3}} \right)dx} } \right] + C$

Answer 1

Hint:Solve by integrating the given expression in parts: \[\int {f(x)g(x)dx} = f(x)\int {g(x)dx} - \int {\left[
{f'(x)\int {g(x)dx} } \right]dx} \]. Make an appropriate substitution. If we substitute x = f(t), then
\[dx{\text{ }} = {\text{ }}f'\left( t \right){\text{ }}dt\]and $\int {f(x)dx} = \int {f\left[ {f(t)} \right]f'(t)dt} $.
Substitute ${x^3} = t$ and integrate by parts.
Recall that $\dfrac{d}{{dx}}{x^n} = n{x^{n - 1}}$ and that if $\int {f(x)dx} = g(x)$, then $\int {f(y)dy} =
g(y)$.

Complete step by step Solution:
Let's say $I = \int {{x^5}f\left( {{x^3}} \right)dx} $, which can also be written as:
$ \Rightarrow I = \int {{x^3}.{x^2}f\left( {{x^3}} \right)dx} $

Substituting ${x^3} = t$, we have $3{x^2}dx = dt$.
And, $I = \dfrac{1}{3}\int {tf(t)dt} $.

Integrating by parts, taking t as the first function and f(t) as the second function, we get:
$ \Rightarrow I = \dfrac{1}{3}\left[ {t\int {f(t)dt} - \int {\left( {\dfrac{d}{{dt}}t\int {f(t)dt} } \right)dt} }
\right] + C$

Differentiating ‘t’ w.r.t. ‘t’, we get ‘1’; so
$ \Rightarrow I = \dfrac{1}{3}\left[ {t\int {f(t)dt} - \int {\left( {\int {f(t)dt} } \right)dt} } \right] + C$
It is given that $\int {f(x)dx} = \Psi (x)$

On Substituting value in equation, we get;
$ \Rightarrow I = \dfrac{1}{3}\left[ {t\Psi (t) - \int {\Psi (t)dt} } \right] + C$
Back substituting ${x^3} = t$, we have $3{x^2}dx = dt$, we get:

$ \Rightarrow I = \dfrac{1}{3}\left[ {{x^3}\Psi ({x^3}) - \int {\Psi ({x^3})\left( {3{x^2}} \right)dx} } \right] +
C$
$ \Rightarrow I = \dfrac{1}{3}{x^3}\Psi ({x^3}) - \int {{x^2}\Psi ({x^3})dx} + C$
Hence, the correct answer is B. $\dfrac{1}{3}{x^3}\Psi ({x^3}) - \int {{x^2}\Psi ({x^3})dx} + C$.

Note:integration by parts or partial integration is a process that finds the integral of a product of functions in terms of the integral of the product of their derivative and antiderivative. When integrating by parts:

\[\int {f(x)g(x)dx} \] = \[f(x)\int {g(x)dx} \] − \[\int {\left[ {f'(x)\int {g(x)dx} } \right]dx} \], make sure to select the functions f(x) and g(x) in such a manner that we can integrate the derivative of the first function f(x) easily. Usually, it's better to follow the rule of ILATE when selecting the first and the second functions.
I: Inverse Trigonometric Functions
L: Logarithmic Functions
A: Algebraic Functions
T: Trigonometric Functions
E: Exponential Functions

Application of Integrals are very important. Integrals are used to find the area under the curve and we can solve differential equations using integration.