$x^2$ 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.

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.

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

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

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

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

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.

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

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

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

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

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

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

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

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.

Latest Math Topics

Aug 31, 2024

Aug 07, 2024

Jul 24, 2024

Dec 13, 2023

Latest Math Problems

Sep 04, 2024

Jan 30, 2024

Oct 15, 2023

Copyright © 2012 - 2023 Math Doubts, All Rights Reserved