Math Doubts

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.

integration by parts question problem

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 $x$.

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

Math Doubts

A best free mathematics education website for students, teachers and researchers.

Maths Topics

Learn each topic of the mathematics easily with understandable proofs and visual animation graphics.

Maths Problems

Learn how to solve the math problems in different methods with understandable steps and worksheets on every concept for your practice.

Learn solutions

Subscribe us

You can get the latest updates from us by following to our official page of Math Doubts in one of your favourite social media sites.

Copyright © 2012 - 2022 Math Doubts, All Rights Reserved