Evaluate $\displaystyle \int{x\sin{x}}\,dx$

The indefinite integral of the product of $x$ and sine of angle $x$ should have to be evaluated with respect to $x$ in this indefinite integration question.

In this indefinite integration problem, it is given that two functions $x$ and $\sin{x}$ are multiplied to form a function by their product. The exponent of the variable $x$ can be reduced by the power rule of derivatives. Similarly, the integral of sine function can also be calculated. Therefore, the indefinite integral of the product of $x$ and $\sin{x}$ can be calculated with respect to $x$ by the integration by parts formula.

Find Derivative and Antiderivative of functions

Assume that $u \,=\, x$ and $dv \,=\, \sin{x}\,dx$ by the change of variables technique.

Firstly, differentiate the both sides of the equation $u \,=\, x$ with respect to $x$ to evaluate the differential element $du$.

$\implies$ $\dfrac{d}{dx}{(u)} \,=\, \dfrac{d}{dx}{(x)}$

$\implies$ $\dfrac{du}{dx} \,=\, \dfrac{dx}{dx}$

$\implies$ $\dfrac{du}{dx} \,=\, 1$

$\implies$ $du \,=\, 1 \times dx$

$\,\,\,\therefore\,\,\,\,\,\,$ $du \,=\, dx$

Now, integrate the both sides of the equation $dv \,= \sin{x}\,dx$ to find the variable $v$.

$\implies$ $\displaystyle \int{}dv \,=\, \int{\sin{x}}\,dx$

$\implies$ $\displaystyle \int{1} \times dv \,=\, \int{\sin{x}}\,dx$

According to the integral of one rule, the derivative of one with respect to $v$ is $v$, and the integral of sine of angle $x$ with respect to $x$ is negative of cosine of angle $x$ as per the integral rule of sine function.

$\,\,\,\therefore\,\,\,\,\,\,$ $v \,=\, -\cos{x}$

Expand Indefinite integral by Integration by parts

Now, write the integration by parts rule in mathematical form.

$\displaystyle \int{u}\,dv$ $\,=\,$ $uv$ $-$ $\displaystyle \int{v}\,du$

We have known already that

$(1).\,\,$ $u \,=\, x$

$(2).\,\,$ $v \,=\, -\cos{x}$

$(3).\,\,$ $du \,=\, dx$

$(4).\,\,$ $dv \,=\, \sin{x}\,dx$

It is time to substitute them in the integration by parts formula to expand the integral of the product of $x$ and $\sin{x}$.

$\implies$ $\displaystyle \int{x\sin{x}}\,dx$ $\,=\,$ $x(-\cos{x})$ $-$ $\displaystyle \int{(-\cos{x})}\,dx$

Evaluate the Indefinite integral of the function

The indefinite integral of the product of $x$ and $\sin{x}$ functions with respect to $x$ can be calculated by evaluating its expansion. So, let us concentrate on every part in the expression on the right hand side of the equation.

$\,\,=\,\,$ $x(-\cos{x})$ $-$ $\displaystyle \int{(-\cos{x})}\,dx$

$\,\,=\,\,$ $-x\cos{x}$ $+$ $\displaystyle \int{\cos{x}}\,dx$

According to the integral rule of cos function, the indefinite integral of cosine of angle $x$ can be evaluated with respect to $x$.

$\,\,=\,\,$ $-x\cos{x}$ $+$ $\sin{x}$ $+$ $c$