Question

How do you find the antiderivative of $\int{\left( {{x}^{3}}\sin x \right)}dx$ ?

Answer 1

Hint: We have been given an expression consisting of a 3-degree x-variable and a trigonometric function whose antiderivative is to be found. From our prior knowledge of basic calculus, we know that the antiderivative is another name for the integral of the function. Thus, we shall integrate the given expression by integration by parts. Further, we shall keep using the ILATE rule until the expression is totally simplified.

Complete step by step solution:
Given that $\int{\left( {{x}^{3}}\sin x \right)}dx$.
This can be also written as $\int{\left( {{x}^{3}} \right)\left( \sin x \right)}dx$ as the product of ${{x}^{3}}$ and sine of x function.
$\Rightarrow \int{\left( {{x}^{3}}\sin x \right)}dx=\int{\left( {{x}^{3}} \right)\left( \sin x \right)}dx$
Here, we shall apply integration by parts. Here, our first function is $\sin x$ and our second function is ${{x}^{3}}$.
\[\Rightarrow \int{\left( {{x}^{3}}\sin x \right).dx={{x}^{3}}.\int{\sin x.dx}}-\int{\left( \int{\sin x.dx} \right).\left( \dfrac{d}{dx}{{x}^{3}} \right)}.dx\]
We know by the formulae of differentiation that $\dfrac{d}{dx}{{x}^{n}}=n{{x}^{n-1}}$ and by the formulae of integration that $\int{\sin x.dx=-\cos x+C}$.
\[\Rightarrow \int{\left( {{x}^{3}}\sin x \right).dx={{x}^{3}}.\left( -\cos x \right)}-\int{\left( -\cos x \right).\left( 3{{x}^{2}} \right)}.dx\]
\[\Rightarrow \int{\left( {{x}^{3}}\sin x \right).dx=-{{x}^{3}}\cos x}+3\int{{{x}^{2}}\cos x}.dx\] …………… equation (1)
Since the second term on the right hand side is not simplified yet thus we shall apply integration by parts again on it.
Let ${{I}_{1}}=\int{{{x}^{2}}\cos x}.dx$
Here, our first function is $\cos x$ and our second function is ${{x}^{2}}$.
\[\Rightarrow {{I}_{1}}={{x}^{2}}.\int{\cos x.dx}-\int{\left( \int{\cos x.dx} \right).\left( \dfrac{d}{dx}{{x}^{2}} \right)}.dx\]
We know by the formulae of differentiation that $\dfrac{d}{dx}{{x}^{n}}=n{{x}^{n-1}}$ and by the formulae of integration that $\int{\cos x.dx=\sin x+C}$.
\[\Rightarrow {{I}_{1}}={{x}^{2}}.\sin x-\int{\sin x.\left( 2x \right)}.dx\]
\[\Rightarrow {{I}_{1}}={{x}^{2}}.\sin x-\int{2x\sin x}.dx\]
Substituting the value of ${{I}_{1}}$ in equation (1), we get
\[\Rightarrow \int{\left( {{x}^{3}}\sin x \right).dx=-{{x}^{3}}\cos x}+3\left( {{x}^{2}}.\sin x-\int{2x\sin x}.dx \right)\]
\[\Rightarrow \int{\left( {{x}^{3}}\sin x \right).dx=-{{x}^{3}}\cos x}+3{{x}^{2}}.\sin x-3\int{2x\sin x}.dx\]
\[\Rightarrow \int{\left( {{x}^{3}}\sin x \right).dx=-{{x}^{3}}\cos x}+3{{x}^{2}}.\sin x-6\int{x\sin x}.dx\] ……………… equation (2)
Since the last term on the right hand side is not simplified yet again thus we shall apply integration by parts again on it.
Let ${{I}_{2}}=\int{x\sin x}.dx$
Here, our first function is $\sin x$ and our second function is $x$.
\[\Rightarrow {{I}_{2}}=x.\int{\sin x.dx}-\int{\left( \int{\sin x.dx} \right).\left( \dfrac{d}{dx}x \right)}.dx\]
\[\Rightarrow {{I}_{2}}=x.\cos x-\int{-\cos x.\left( 1 \right)}.dx\]
\[\Rightarrow {{I}_{2}}=x.\cos x+\int{\cos x.}dx\]
We know by the formulae of integration that $\int{\cos x.dx=\sin x+C}$.
\[\Rightarrow {{I}_{2}}=x.\cos x+\sin x+C\]
Substituting the value of ${{I}_{2}}$ in equation (2), we get
\[\Rightarrow \int{\left( {{x}^{3}}\sin x \right).dx=-{{x}^{3}}\cos x}+3{{x}^{2}}.\sin x-6\left( x.\cos x+\sin x+C \right)\]
\[\Rightarrow \int{\left( {{x}^{3}}\sin x \right).dx=-{{x}^{3}}\cos x}+3{{x}^{2}}.\sin x-6x.\cos x+6\sin x+C\]
Therefore, the antiderivative of $\int{\left( {{x}^{3}}\sin x \right)}dx$ is \[-{{x}^{3}}\cos x+3{{x}^{2}}.\sin x-6x.\cos x+6\sin x+C\].

Note: Since we do not have a direct formula of integration for the given expression which is a product of two different functions, therefore, we integrated this function by applying integration by parts. Here, the function was manipulated by again and again integrating it by parts until the most all the terms are property integrated.