Question

How do you find the antiderivative of $\int{{{x}^{2}}\cos xdx}$.

Answer 1

Hint: In the problem we have two functions which are in multiplication one is ${{x}^{2}}$ and the second one is $\cos x$. In integration we have the $uv$ formula as $\int{uv}=u\int{v}-\int{\left( {{u}^{'}} \right)\int{v}}$. So, we will use the ILATE rule and determine the values of $u$ and $v$. After getting these values we will use the $uv$ formula and do simplification. Now we will get another equation which is in the same form. So, we will again use the $uv$ formula and simplify the obtained equation. Now we will get the required result.

Complete step by step answer:
Given that, $\int{{{x}^{2}}\cos xdx}$.
In the above equation we have two functions, one is ${{x}^{2}}$ which is an algebraic function and the second one is $\cos x$ which is a trigonometric function. By using the ILATE formula
$u={{x}^{2}}$, $v=\cos x$.
Applying $uv$ formula in the given equation, then we will get
$\int{{{x}^{2}}\cos xdx}={{x}^{2}}\int{\cos xdx}-\int{\left( {{\left( {{x}^{2}} \right)}^{'}}\int{\cos xdx} \right)dx}$
We know that ${{\left( {{x}^{2}} \right)}^{'}}=2x$, $\int{\cos xdx}=\sin x+C$, then we will get
$\begin{align}
  & \Rightarrow \int{{{x}^{2}}\cos xdx}={{x}^{2}}\left( \sin x \right)-\int{2x\sin xdx} \\
 & \Rightarrow \int{{{x}^{2}}\cos xdx}={{x}^{2}}\sin x-2\int{x\sin xdx}....\left( \text{i} \right) \\
\end{align}$
In the above equation, we have $\int{x\sin xdx}$ which is similar to the $uv$ rule. So again, using the ILATE formula then we will get
$u=x$, $v=\sin x$.
Now the value of $\int{x\sin xdx}$ is given by
$\int{x\sin xdx}=x\int{\sin x}-\int{\left( {{\left( x \right)}^{'}}\int{\sin xdx} \right)dx}$
We have ${{x}^{'}}=1$, $\int{\sin xdx}=-\cos x+C$, then we will get
$\begin{align}
  & \int{x\sin xdx}=-x\cos x-\int{\left( -\cos x \right)dx} \\
 & \Rightarrow \int{x\sin xdx}=-x\cos x+\int{\cos xdx} \\
\end{align}$
We have $\int{\cos xdx}=\sin x+C$, then we will get
$\Rightarrow \int{x\sin xdx}=-x\cos x+\sin x+C$
Substituting the above value in the equation $\left( \text{i} \right)$, then we will get
$\Rightarrow \int{{{x}^{2}}\cos xdx}={{x}^{2}}\sin x-2\left( -x\cos x+\sin x \right)+C$
Simplifying the above equation by applying multiplication distribution law, then we will get
$\Rightarrow \int{{{x}^{2}}\cos xdx}={{x}^{2}}\sin x+2x\cos x-2\sin x+C$
Taking $\sin x$ common from the terms ${{x}^{2}}\sin x-2\sin x$, then we will get
$\Rightarrow \int{{{x}^{2}}\cos xdx}=\left( {{x}^{2}}-2 \right)\sin x+2x\cos x+C$

Hence the value of $\int{{{x}^{2}}\cos xdx}$ is $\left( {{x}^{2}}-2 \right)\sin x+2x\cos x+C$.

Note: In the above problem we have used the $uv$, ILATE formulas several times. We need to follow some rules while using the $uv$ rule which is denoted by ILATE which indicates the order of giving priority for a function. It states that the order of the priority of functions as Inverse, Logarithmic, Algebraic, Trigonometric, Exponential. From the above priority table, we will choose the functions $u$ and $v$ in $uv$ rule.