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Hint: Assume the integral to be a quantity I. Then perform the integration by parts where the integrand is expressed as the product of two functions of x in the form $\int{\text{u}\cdot \text{vd}x}$, where u and v are two differentiable functions of x. The general formula for integrating an integral of form $\int{\text{u}\cdot \text{vd}x}$is,

\[\int{u\cdot v\text{d}x}\text{ = }u\int{v\text{d}x\text{ }-\text{ }\int{\left[ \dfrac{du}{dx}\cdot \int{vdx} \right]}\text{d}x}\]

__Complete step-by-step answer:__

Let us assume, I = $\int{x\sin x\cos xdx}$

Now, let us manipulate the integrand so that it can be expressed in a much simpler form.

We know, $\sin 2x\text{ = 2}\sin x\cos x$

Therefore, we can say, $\sin x\cos x\text{ = }\dfrac{\sin 2x}{2}$

Thus, the integration can be expressed as, I = $\dfrac{1}{2}\int{x\sin 2xdx}$

Let, $\text{I }\!\!'\!\!\text{ = }\int{x\sin 2xdx}$. Thus, $\text{I = }\dfrac{\text{I }\!\!'\!\!\text{ }}{2}$.

Now, observe that the integral $\text{I }\!\!'\!\!\text{ }$ is of the form $\int{\text{u}\cdot \text{vd}x}$, where u = x and v = sin 2x. Also, both u and v are differentiable functions of x. Thus, we can apply the process of integration by parts to find the value of $\text{I }\!\!'\!\!\text{ }$, and subsequently I.

\[\begin{align}

& \int{x\sin 2xdx} \\

& =\text{ }x\int{\sin 2xdx\text{ }-\text{ }\int{\left[ \dfrac{d\left( x \right)}{dx}\cdot \int{\sin 2xdx} \right]}}dx \\

& =\text{ }x\cdot \left( \dfrac{-\cos 2x}{2} \right)\text{ }-\text{ }\int{\left[ 1\cdot \left( \dfrac{-\cos 2x}{2} \right) \right]dx}\text{ }\left( \because \text{ }\int{\sin axdx}\text{ = }\dfrac{-\cos ax}{a} \right) \\

& =\text{ }\dfrac{-x\cos 2x}{2}\text{ + }\int{\dfrac{\cos 2x}{2}dx} \\

& =\text{ }\dfrac{-x\cos 2x}{2}\text{ + }\dfrac{\sin 2x}{4}\text{ + c }\left( \because \text{ }\int{\cos axdx}\text{ = }\dfrac{sinax}{a} \right) \\

\end{align}\]

where, c is the constant of integration.

Thus, we find the value of $\text{I }\!\!'\!\!\text{ }$ to be \[\dfrac{-x\cos 2x}{2}\text{ + }\dfrac{\sin 2x}{4}\text{ + c}\].

We know that, $\text{I = }\dfrac{\text{I }\!\!'\!\!\text{ }}{2}$

Therefore, the value of I

$\begin{align}

& =\text{ }\dfrac{1}{2}\cdot \left( \dfrac{-x\cos 2x}{2}\text{ + }\dfrac{\sin 2x}{4}\text{ + c} \right) \\

& =\text{ }\dfrac{-x\cos 2x}{4}\text{ + }\dfrac{\sin 2x}{8}\text{ + k} \\

\end{align}$

where, k is the constant of integration.

Note: i) While choosing u and v as functions of x for integration by parts, we need to select in such a way that $\int{vdx}$ and $\int{\left[ \dfrac{du}{dx}\cdot \int{vdx} \right]dx}$ are simple to integrate. For example, in this problem, if we choose u = xsinx and v = cosx before manipulating the integrand in an easier way, then it would have been much harder to integrate $\int{\left[ \dfrac{du}{dx}\cdot \int{vdx} \right]dx}$.

ii) u and v are generally chosen in such a way that these functions are kept in order of ILATE, where, I is Inverse function, L is logarithmic function, A is algebraic function, T is trigonometric function and E is exponential function.

\[\int{u\cdot v\text{d}x}\text{ = }u\int{v\text{d}x\text{ }-\text{ }\int{\left[ \dfrac{du}{dx}\cdot \int{vdx} \right]}\text{d}x}\]

Let us assume, I = $\int{x\sin x\cos xdx}$

Now, let us manipulate the integrand so that it can be expressed in a much simpler form.

We know, $\sin 2x\text{ = 2}\sin x\cos x$

Therefore, we can say, $\sin x\cos x\text{ = }\dfrac{\sin 2x}{2}$

Thus, the integration can be expressed as, I = $\dfrac{1}{2}\int{x\sin 2xdx}$

Let, $\text{I }\!\!'\!\!\text{ = }\int{x\sin 2xdx}$. Thus, $\text{I = }\dfrac{\text{I }\!\!'\!\!\text{ }}{2}$.

Now, observe that the integral $\text{I }\!\!'\!\!\text{ }$ is of the form $\int{\text{u}\cdot \text{vd}x}$, where u = x and v = sin 2x. Also, both u and v are differentiable functions of x. Thus, we can apply the process of integration by parts to find the value of $\text{I }\!\!'\!\!\text{ }$, and subsequently I.

\[\begin{align}

& \int{x\sin 2xdx} \\

& =\text{ }x\int{\sin 2xdx\text{ }-\text{ }\int{\left[ \dfrac{d\left( x \right)}{dx}\cdot \int{\sin 2xdx} \right]}}dx \\

& =\text{ }x\cdot \left( \dfrac{-\cos 2x}{2} \right)\text{ }-\text{ }\int{\left[ 1\cdot \left( \dfrac{-\cos 2x}{2} \right) \right]dx}\text{ }\left( \because \text{ }\int{\sin axdx}\text{ = }\dfrac{-\cos ax}{a} \right) \\

& =\text{ }\dfrac{-x\cos 2x}{2}\text{ + }\int{\dfrac{\cos 2x}{2}dx} \\

& =\text{ }\dfrac{-x\cos 2x}{2}\text{ + }\dfrac{\sin 2x}{4}\text{ + c }\left( \because \text{ }\int{\cos axdx}\text{ = }\dfrac{sinax}{a} \right) \\

\end{align}\]

where, c is the constant of integration.

Thus, we find the value of $\text{I }\!\!'\!\!\text{ }$ to be \[\dfrac{-x\cos 2x}{2}\text{ + }\dfrac{\sin 2x}{4}\text{ + c}\].

We know that, $\text{I = }\dfrac{\text{I }\!\!'\!\!\text{ }}{2}$

Therefore, the value of I

$\begin{align}

& =\text{ }\dfrac{1}{2}\cdot \left( \dfrac{-x\cos 2x}{2}\text{ + }\dfrac{\sin 2x}{4}\text{ + c} \right) \\

& =\text{ }\dfrac{-x\cos 2x}{4}\text{ + }\dfrac{\sin 2x}{8}\text{ + k} \\

\end{align}$

where, k is the constant of integration.

Note: i) While choosing u and v as functions of x for integration by parts, we need to select in such a way that $\int{vdx}$ and $\int{\left[ \dfrac{du}{dx}\cdot \int{vdx} \right]dx}$ are simple to integrate. For example, in this problem, if we choose u = xsinx and v = cosx before manipulating the integrand in an easier way, then it would have been much harder to integrate $\int{\left[ \dfrac{du}{dx}\cdot \int{vdx} \right]dx}$.

ii) u and v are generally chosen in such a way that these functions are kept in order of ILATE, where, I is Inverse function, L is logarithmic function, A is algebraic function, T is trigonometric function and E is exponential function.

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