Question

What is the integral of $\int{x{{\sin }^{2}}\left( x \right)dx}$?

Answer 1

Hint: Use the trigonometric identity ${{\sin }^{2}}x=\dfrac{1-\cos 2x}{2}$ and simplify the function inside the integral. Now, break the integral into two parts and for the first part use the formula $\int{{{x}^{n}}dx}=\dfrac{{{x}^{n+1}}}{n+1}$ to find its integral. For the second part use the ILATE rule and consider x as function 1 $\left( {{f}_{1}}\left( x \right) \right)$ and $\cos 2x$ as function 2 $\left( {{f}_{2}}\left( x \right) \right)$ and apply the rule of integration by parts given as $\int{{{f}_{1}}\left( x \right).{{f}_{2}}\left( x \right)dx}=\left[ {{f}_{1}}\left( x \right)\int{{{f}_{2}}\left( x \right)dx} \right]-\int{\left[ {{f}_{1}}'\left( x \right)\int{{{f}_{2}}\left( x \right)dx} \right]dx}$ to get the answer. Here, \[{{f}_{1}}'\left( x \right)=\dfrac{d}{dx}\left( {{f}_{1}}\left( x \right) \right)\]. Use the formulas \[\int{\cos \left( ax+b \right)dx}=\dfrac{\sin \left( ax+b \right)}{a}\] and \[\int{\sin \left( ax+b \right)dx}=\dfrac{-\cos \left( ax+b \right)}{a}\] to evaluate the integral.

Complete step by step answer:
Here we are asked to find the integral of the function \[x{{\sin }^{2}}\left( x \right)\]. First let us simplify the trigonometric function using the half angle formula. Let us assume the integral as I, so we have,
\[\Rightarrow I=\int{x{{\sin }^{2}}\left( x \right)dx}\]
Using the half angle trigonometric identity given as ${{\sin }^{2}}x=\dfrac{1-\cos 2x}{2}$ we get,
\[\Rightarrow I=\int{x\left( \dfrac{1-\cos 2x}{2} \right)dx}\]
Since, 2 is a constant so it can be taken out of the integral sign, we have,
\[\Rightarrow I=\dfrac{1}{2}\int{\left( x-x\cos 2x \right)dx}\]
Breaking the integral into two parts we get,
\[\Rightarrow I=\dfrac{1}{2}\left( \int{xdx}-\int{x\cos 2xdx} \right)\]
Using the formula $\int{{{x}^{n}}dx}=\dfrac{{{x}^{n+1}}}{n+1}$ for the integral of the first term we get,
\[\begin{align}
  & \Rightarrow I=\dfrac{1}{2}\left( \dfrac{{{x}^{1+1}}}{1+1}-\int{x\cos 2xdx} \right) \\
 & \Rightarrow I=\dfrac{1}{2}\left( \dfrac{{{x}^{2}}}{2}-\int{x\cos 2xdx} \right) \\
\end{align}\]
Clearly we can see that in the second part we have a product of an algebraic function and a trigonometric function so we need to apply integration by parts to find the integral. Now, according to the ILATE rule we have to assume x as the function 1 $\left( {{f}_{1}}\left( x \right) \right)$ and $\cos 2x$ as function 2 $\left( {{f}_{2}}\left( x \right) \right)$. Here, ILATE stands for: -
I – Inverse trigonometric function
L – Logarithmic function
A – Algebraic function
T – Trigonometric function
E – Exponential function
The numbering of the functions is done according to the order of appearance in the above list. Therefore assuming x as function 1 $\left( {{f}_{1}}\left( x \right) \right)$ and $\cos 2x$ as function 2 $\left( {{f}_{2}}\left( x \right) \right)$we have the formula $\int{{{f}_{1}}\left( x \right).{{f}_{2}}\left( x \right)dx}=\left[ {{f}_{1}}\left( x \right)\int{{{f}_{2}}\left( x \right)dx} \right]-\int{\left[ {{f}_{1}}'\left( x \right)\int{{{f}_{2}}\left( x \right)dx} \right]dx}$ to calculate the integral of product of two functions. So we get,
\[\begin{align}
  & \Rightarrow I=\dfrac{1}{2}\left[ \dfrac{{{x}^{2}}}{2}-\left\{ x\int{\cos 2xdx}-\int{\left( \left( \int{\cos 2xdx} \right)\times \dfrac{d\left( x \right)}{dx} \right)dx} \right\} \right] \\
 & \Rightarrow I=\dfrac{1}{2}\left[ \dfrac{{{x}^{2}}}{2}-\left\{ x\int{\cos 2xdx}-\int{\left( \left( \int{\cos 2xdx} \right) \right)dx} \right\} \right] \\
\end{align}\]
Using the formula \[\int{\cos \left( ax+b \right)dx}=\frac{\sin \left( ax+b \right)}{a}\] we get,
\[\begin{align}
  & \Rightarrow I=\dfrac{1}{2}\left[ \dfrac{{{x}^{2}}}{2}-\left\{ x\left( \dfrac{\sin 2x}{2} \right)-\int{\dfrac{\sin 2x}{2}dx} \right\} \right] \\
 & \Rightarrow I=\dfrac{1}{2}\left[ \dfrac{{{x}^{2}}}{2}-\left\{ x\left( \dfrac{\sin 2x}{2} \right)-\dfrac{1}{2}\int{\sin 2xdx} \right\} \right] \\
\end{align}\]
Using the formula \[\int{\sin \left( ax+b \right)dx}=\dfrac{-\cos \left( ax+b \right)}{a}\] we get,
\[\begin{align}
  & \Rightarrow I=\dfrac{1}{2}\left[ \dfrac{{{x}^{2}}}{2}-\left\{ x\left( \dfrac{\sin 2x}{2} \right)-\dfrac{1}{2}\left( \dfrac{-\cos 2x}{2} \right) \right\} \right] \\
 & \therefore I=\dfrac{1}{2}\left[ \dfrac{{{x}^{2}}}{2}+\dfrac{x\sin 2x}{2}-\dfrac{\cos 2x}{4} \right]+c \\
\end{align}\]
Here ‘c’ is the constant of integration as we are evaluating an indefinite integral.

Note: Note that we cannot apply the ILATE rule and integration by parts directly in the function \[x{{\sin }^{2}}\left( x \right)\] because we don’t have any direct formula for the integration of the function ${{\sin }^{2}}x$ so it must be converted into $\cos 2x$ using the half angle formula. Also, remember that you cannot apply the formula $\int{{{x}^{n}}dx}=\dfrac{{{x}^{n+1}}}{n+1}$ for n = -1 because in that case we have the function $\dfrac{1}{x}$ whose integral is $\ln x$.