Math Doubts

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.

Math Doubts
Math Doubts is a best place to learn mathematics and from basics to advanced scientific level for students, teachers and researchers. Know more
Follow us on Social Media
Math Problems

Learn how to solve easy to difficult mathematics problems of all topics in various methods with step by step process and also maths questions for practising.

Learn more