Question

How do you integrate $\int{{{x}^{2}}{{\sin }^{2}}xdx}$ using integration by parts?

Answer 1

Hint: To integrate the given integration using integration by parts, we are going to use the following integration by parts formula which is equal to: $\int{f\left( x \right)g\left( x \right)dx}=f\left( x \right)\int{g\left( x \right)dx}-\int{f'\left( x \right)}\int{g\left( x \right)dx}$. The catch in this integration is that what could be the first term and what could be the second term. And this priority of the function is decided by the mnemonics named “ILATE” and we will discuss each letter in the below solution.

Complete step-by-step answer:
The integration which we have to find is as follows:
$\int{{{x}^{2}}{{\sin }^{2}}xdx}$
Now, we are going to integrate by using integration by parts. The formula for integration by parts is as follows:
$\int{f\left( x \right)g\left( x \right)dx}=f\left( x \right)\int{g\left( x \right)dx}-\int{f'\left( x \right)}\int{g\left( x \right)dx}$ …………… (1)
In the above formula, $f\left( x \right)\And g\left( x \right)$ are two functions but here, we should know which function will get the priority over another. To decide the priority there is mnemonic which is equal to:
ILATE
Now, in the above mnemonic, “I” represents inverse functions, “L” represents logarithmic functions, “A” represents algebraic functions, “T” represents trigonometric functions and “E” represents exponential functions. Now, the order of the letters in this word “ILATE” will be the order of priority for the functions.
So, using this mnemonic in the given integration, in the given integration, we have two functions, algebraic and trigonometric functions so algebraic will get priority over trigonometric functions. Now, $f\left( x \right)={{x}^{2}}\And g\left( x \right)={{\sin }^{2}}x$ then substituting these values of f(x) and g(x) in eq. (1) we get,
$\int{{{x}^{2}}{{\sin }^{2}}xdx}={{x}^{2}}\int{{{\sin }^{2}}xdx}-\int{\left( {{x}^{2}} \right)'}\int{{{\sin }^{2}}xdx}$ ……………. (2)
The integration of ${{\sin }^{2}}x$ with respect to x is as follows:
$\begin{align}
  & \int{{{\sin }^{2}}xdx}=\dfrac{1}{2}\int{\left( 1-\cos 2x \right)}dx \\
 & \Rightarrow \int{{{\sin }^{2}}xdx}=\dfrac{1}{2}\int{dx}-\dfrac{1}{2}\int{\cos 2xdx} \\
\end{align}$
Also, we know that integration of $\cos 2x$ with respect to x we get,
$\int{\cos 2xdx}=\dfrac{1}{2}\sin 2x$
Using the above relation in the above integration we get,
$\begin{align}
  & \Rightarrow \int{{{\sin }^{2}}xdx}=\dfrac{1}{2}x-\dfrac{1}{2}\times \dfrac{1}{2}\sin 2x \\
 & \Rightarrow \int{{{\sin }^{2}}xdx}=\dfrac{1}{2}x-\dfrac{1}{4}\sin 2x \\
\end{align}$
And the differentiation of ${{x}^{2}}$ with respect to x we get,
$\left( {{x}^{2}} \right)'=2x$
Using the above integration and differentiation in eq. (2) we get,
$\begin{align}
  & \int{{{x}^{2}}{{\sin }^{2}}xdx}={{x}^{2}}\left( \dfrac{1}{2}x-\dfrac{1}{4}\sin 2x \right)-\int{\left( 2x \right)}\left( \dfrac{1}{2}x-\dfrac{1}{4}\sin 2x \right)dx \\
 & \Rightarrow \int{{{x}^{2}}{{\sin }^{2}}xdx}=\dfrac{{{x}^{3}}}{2}-\dfrac{1}{4}{{x}^{2}}\sin 2x-\left( \int{{{x}^{2}}-\dfrac{1}{2}x\sin 2x} \right)dx \\
 & \Rightarrow \int{{{x}^{2}}{{\sin }^{2}}xdx}=\dfrac{{{x}^{3}}}{2}-\dfrac{1}{4}{{x}^{2}}\sin 2x-\int{{{x}^{2}}dx}+\dfrac{1}{2}\int{x\sin 2xdx} \\
 & \Rightarrow \int{{{x}^{2}}{{\sin }^{2}}xdx}=\dfrac{{{x}^{3}}}{2}-\dfrac{1}{4}{{x}^{2}}\sin 2x-\dfrac{{{x}^{3}}}{3}+\dfrac{1}{2}\int{x\sin 2xdx}.........(3) \\
\end{align}$
We are going to do the integration of $x\sin 2x$ with respect to x by using integration by parts we get,
\[\begin{align}
  & \int{x\sin 2xdx}=x\int{\sin 2xdx}-\int{x'}\int{\sin 2xdx} \\
 & \Rightarrow \int{x\sin 2xdx}=-x\dfrac{\cos 2x}{2}-\int{1\left( -\dfrac{1}{2}\cos 2x \right)} \\
 & \Rightarrow \int{x\sin 2xdx}=-x\dfrac{\cos 2x}{2}+\dfrac{1}{2}\int{\cos 2xdx} \\
\end{align}\]
\[\begin{align}
  & \Rightarrow \int{x\sin 2xdx}=-x\dfrac{\cos 2x}{2}+\dfrac{1}{2}\left( \dfrac{1}{2}\sin 2x \right) \\
 & \Rightarrow \int{x\sin 2xdx}=-x\dfrac{\cos 2x}{2}+\dfrac{1}{4}\sin 2x \\
\end{align}\]
Substituting the above integration in eq. (3) we get,
$\begin{align}
  & \Rightarrow \int{{{x}^{2}}{{\sin }^{2}}xdx}=\dfrac{{{x}^{3}}}{2}-\dfrac{1}{4}{{x}^{2}}\sin 2x-\dfrac{{{x}^{3}}}{3}+\dfrac{1}{2}\left( -x\dfrac{\cos 2x}{2}+\dfrac{1}{4}\sin 2x \right) \\
 & \Rightarrow \int{{{x}^{2}}{{\sin }^{2}}xdx}=\dfrac{{{x}^{3}}}{2}-\dfrac{1}{4}{{x}^{2}}\sin 2x-\dfrac{{{x}^{3}}}{3}-\dfrac{x\cos 2x}{4}+\dfrac{1}{8}\sin 2x \\
\end{align}$
Solving the same power of x in the R.H.S of the above equation we get,
$\begin{align}
  & \Rightarrow \int{{{x}^{2}}{{\sin }^{2}}xdx}={{x}^{3}}\left( \dfrac{1}{2}-\dfrac{1}{3} \right)+\dfrac{1}{4}\sin 2x\left( \dfrac{1}{2}-{{x}^{2}} \right)-\dfrac{x\cos 2x}{4} \\
 & \Rightarrow \int{{{x}^{2}}{{\sin }^{2}}xdx}={{x}^{3}}\left( \dfrac{3-2}{6} \right)+\dfrac{1}{4}\sin 2x\left( \dfrac{1}{2}-{{x}^{2}} \right)-\dfrac{x\cos 2x}{4} \\
 & \Rightarrow \int{{{x}^{2}}{{\sin }^{2}}xdx}={{x}^{3}}\left( \dfrac{1}{6} \right)+\dfrac{1}{4}\sin 2x\left( \dfrac{1}{2}-{{x}^{2}} \right)-\dfrac{x\cos 2x}{4} \\
\end{align}$
Hence, we have done the given integration using integration by parts as follows:
${{x}^{3}}\left( \dfrac{1}{6} \right)+\dfrac{1}{4}\sin 2x\left( \dfrac{1}{2}-{{x}^{2}} \right)-\dfrac{x\cos 2x}{4}$

Note: The point of making a mistake is in integration of $\sin x\And \cos x$ and the mistake is in putting the sign before the result of integration.
In the integration of $\sin x$ with respect to x we get negative signs in the result whereas in the integration of $\cos x$ with respect to x we get positive signs in the result.
$\begin{align}
  & \int{\sin xdx}=-\cos x \\
 & \int{\cos xdx}=\sin x \\
\end{align}$