Question

How do you integrate the given function $x.{{\cos }^{2}}x$?

Answer 1

Hint: We start solving the problem by equating the given indefinite integral to a variable. We then make use of the result that ${{\cos }^{2}}x=\dfrac{1+\cos 2x}{2}$ to proceed through the problem. We then recall the integration by parts as $\int{f\left( x \right)\times g\left( x \right)dx}=f\left( x \right)\int{g\left( x \right)dx}-\int{\left( \dfrac{d\left( f\left( x \right) \right)}{dx}\int{g\left( x \right)dx} \right)dx}$ to proceed through the problem. We then make the necessary calculations and make use of the results that $\dfrac{d\left( x \right)}{dx}=1$, $\int{adx}=ax+C$ and $\int{\cos axdx}=\dfrac{\sin ax}{a}+C$ to proceed further through the problem. We then make use of the results that $\int{axdx}=\dfrac{a{{x}^{2}}}{2}+C$, $\int{\sin axdx}=\dfrac{-\cos ax}{a}+C$ and then make the necessary calculations to get the required answer.

Complete step by step answer:
According to the problem, we are asked to find the result of the given indefinite integral \[\int{x.{{\cos }^{2}}xdx}\].
Let us assume $I=\int{x.{{\cos }^{2}}xdx}$ ---(1).
We know that ${{\cos }^{2}}x=\dfrac{1+\cos 2x}{2}$. Let us use this result in equation (1).
$\Rightarrow I=\int{x.\left( \dfrac{1+\cos 2x}{2} \right)dx}$ ---(2).
We can see that the integrand is in the form of $\int{f\left( x \right)\times g\left( x \right)dx}$. From integration by parts, we know that $\int{f\left( x \right)\times g\left( x \right)dx}=f\left( x \right)\int{g\left( x \right)dx}-\int{\left( \dfrac{d\left( f\left( x \right) \right)}{dx}\int{g\left( x \right)dx} \right)dx}$. Let us use this result in equation (2).
$\Rightarrow I=x\int{\left( \dfrac{1+\cos 2x}{2} \right)dx}-\int{\left( \dfrac{d}{dx}\left( x \right)\int{\left( \dfrac{1+\cos 2x}{2} \right)dx} \right)dx}$ ---(3).
We know that $\dfrac{d\left( x \right)}{dx}=1$. Let us use this result in equation (3).
$\Rightarrow I=\dfrac{x}{2}\int{\left( 1+\cos 2x \right)dx}-\int{\left( \dfrac{1}{2}\int{\left( 1+\cos 2x \right)dx} \right)dx}$ ---(4).
We know that $\int{adx}=ax+C$, $\int{\cos axdx}=\dfrac{\sin ax}{a}+C$. Let us use these results in equation (4).
\[\Rightarrow I=\dfrac{x}{2}\left( x+\dfrac{\sin 2x}{2} \right)-\int{\left( \dfrac{1}{2}\left( x+\dfrac{\sin 2x}{2} \right) \right)dx}\].
\[\Rightarrow I=\dfrac{x}{2}\left( x+\dfrac{\sin 2x}{2} \right)-\int{\left( \dfrac{1}{2}\left( \dfrac{2x+\sin 2x}{2} \right) \right)dx}\].
\[\Rightarrow I=\dfrac{x}{2}\left( x+\dfrac{\sin 2x}{2} \right)-\dfrac{1}{4}\int{\left( 2x+\sin 2x \right)dx}\] ---(5).
We know that $\int{axdx}=\dfrac{a{{x}^{2}}}{2}+C$, $\int{\sin axdx}=\dfrac{-\cos ax}{a}+C$. Let us use these results in equation (5).
\[\Rightarrow I=\dfrac{x}{2}\left( x+\dfrac{\sin 2x}{2} \right)-\dfrac{1}{4}\left( \dfrac{2{{x}^{2}}}{2}-\dfrac{\cos 2x}{2} \right)+C\].
\[\Rightarrow I=\dfrac{{{x}^{2}}}{2}+\dfrac{x\sin 2x}{4}-\dfrac{{{x}^{2}}}{4}+\dfrac{\cos 2x}{8}+C\].
\[\Rightarrow I=\dfrac{{{x}^{2}}}{4}+\dfrac{x\sin 2x}{4}+\dfrac{\cos 2x}{8}+C\].
$\therefore $ We have found the result of integration of the function $x.{{\cos }^{2}}x$ as \[\dfrac{{{x}^{2}}}{4}+\dfrac{x\sin 2x}{4}+\dfrac{\cos 2x}{8}+C\].

Note:
We can see that the given problem contains a huge amount of calculation, so we need to perform each step carefully to avoid confusion and calculation mistakes. We should not forget to add constant integration while solving this type of problem as this is the common mistake done by students. Similarly, we can expect problems to find the value of the definite integral $\int\limits_{0}^{\dfrac{\pi }{2}}{x.{{\cos }^{2}}xdx}$.