# Evaluate $\displaystyle \int{x^2\sin{x} \,} dx$

$x^2$ is an algebraic function and $\sin{x}$ is a trigonometric function. The two different functions are involved in multiplication and formed an algebraic trigonometric function $x^2\sin{x}$ by their product. In this integral problem, we have to calculate the integration of the function $x^2\sin{x}$ with respect to $x$.

$\displaystyle \int{x^2\sin{x} \,} dx$

In this calculus problem, the functions $x^2$ and $\sin{x}$ are multiplying each other. So, it can be evaluated with respect to $x$ by the integration by parts formula.

### Integration using Integration by Parts

Compare the integration of the product of the functions with integration by parts formula.

$\displaystyle \int{x^2\sin{x} \,} dx$ $\,=\,$ $\displaystyle \int{u}dv$

Take $u = x^2$ and $dv = \sin{x}dx$.

$(1) \,\,\,$ Find $du$ by differentiating the equation $u = x^2$ as per power rule of differentiation.

$\dfrac{d}{dx}{\, (u)} = \dfrac{d}{dx}{\, x^2}$

$\implies$ $\dfrac{du}{dx} = 2x^{2-1}$

$\implies$ $\dfrac{du}{dx} = 2x^1$

$\implies$ $\dfrac{du}{dx} = 2x$

$\,\,\, \therefore \,\,\,\,\,\,$ $du = {2x}dx$

$(2) \,\,\,$ Now, find $v$ by integrating both sides of the equation $dv = \sin{x}dx$. It can be evaluated by the integral rule of sin function.

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

$\implies$ $v+c = -\cos{x}+c$

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

Now, find the integral of the function $x^2\sin{x}$ with respect to $x$ by using integration by parts formula.

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

$\implies$ $\displaystyle \int{x^2\sin{x}}dx$ $\,=\,$ $x^2 \times (-\cos{x})-\displaystyle \int{(-\cos{x})}(2xdx)$

$\implies$ $\displaystyle \int{x^2\sin{x}}dx$ $\,=\,$ $-x^2\cos{x}-\displaystyle \int{(-2x\cos{x})}dx$

$\,\,\, \therefore \,\,\,\,\,\,$ $\displaystyle \int{x^2\sin{x}}dx$ $\,=\,$ $-x^2\cos{x}+2\displaystyle \int{x\cos{x}}dx$

### Continue the integration using Integration by Parts

The indefinite integration of the function $x^2\sin{x}$ is not done completely but it can be finished by applying integration by parts to the function $x\cos{x}$.

$\displaystyle \int{x\cos{x}}dx$ $\,=\,$ $\displaystyle \int{u dv \,} dx$

Take $u = x$ and $dv = \cos{x}dx$.

$(1) \,\,\,$ Find $du$ by differentiating the equation $u = x$ with the derivative of a variable rule.

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

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

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

$(2) \,\,\,$ Now, find $v$ by integrating both sides of the equation $dv = \cos{x}dx$. It can be done by the integral rule of cos function.

$\displaystyle \int{}dv = \int{\cos{x} \,}dx$

$\implies$ $v+c = \sin{x}+c$

$\,\,\, \therefore \,\,\,\,\,\,$ $v = \sin{x}$

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

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

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

$\,\,\, \therefore \,\,\,\,\,\,$ $\displaystyle \int{x\cos{x}}dx$ $\,=\,$ $x\sin{x}+\cos{x}+c$

### Evaluate the Integration of the function

In the first step, the indefinite integration of the function is calculated as follows.

$\displaystyle \int{x^2\sin{x}}dx$ $\,=\,$ $-x^2\cos{x}+2\displaystyle \int{x\cos{x}}dx$

Now, replace the indefinite integration of the function $x\cos{x}$ in the above equation.

$\implies$ $\displaystyle \int{x^2\sin{x}}dx$ $\,=\,$ $-x^2\cos{x}+2(x\sin{x}+\cos{x}+c)$

$\implies$ $\displaystyle \int{x^2\sin{x}}dx$ $\,=\,$ $-x^2\cos{x}+2(x\sin{x}+\cos{x})+2c$

The term $2c$ is a constant term. So, it is simply represented by the integral constant ($c$).

$\,\,\, \therefore \,\,\,\,\,\,$ $\displaystyle \int{x^2\sin{x}}dx$ $\,=\,$ $-x^2\cos{x}+2(x\sin{x}+\cos{x})+c$

Thus, the indefinite integration of the algebraic trigonometric function $x^2\sin{x}$ with respect to $x$ by using integration by parts in integral calculus.

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