Question

How do you integrate $\int {x\sin xdx} $ using integration by parts?

Answer 1

Hint: In this problem, we are given an expression for integration, where the integrand is a combination of algebraic and trigonometric terms. Here, we are asked to integrate the given problem by using integration by parts. By using integration by parts, we are going to expand the differential of a product of functions.

Formula used: Integration by parts $\int {udv = uv - \int {vdu} } $ , where $u$ is a function of $u\left( x \right)$ , $dv$ is a variable $dv$ , $v$ is the function of $v\left( x \right)$ and $du$ is a variable $du$ .

Complete step-by-step solution:
Given integration is $\int {x\sin xdx} $ .
Take $I = \int {x\sin xdx} $ .
We want to integrate by parts, taking $u = x$ , $dv = \sin x$ .
Differentiating $u = x$ we get, $du = dx$ and integrating $dv = \sin x$ we get, $v = - \cos x$ .
$ \Rightarrow \int {udv = uv - \int {vdu} } $
Now, let’s substitute all the values in the formula, we get,
$ \Rightarrow \int {x\sin xdx = - x\cos x + \int {\cos xdx} } $
Now integrating $\cos x$ in the right-hand side, we get,
$ \Rightarrow \int {x\sin xdx = - x\cos x + \sin x + C} $, where C is the constant of integration.
$\therefore ,I = \sin x - x\cos x + C$

Therefore, the given integration by using integration by parts is equal to $\sin x - x\cos x + C$.

Additional Information: Integration by parts is a technique for performing indefinite integration or definite integration by expanding the differential of a product of functions and expressing the original integral in terms of a known integral. Integration by parts is for functions that can be written as the product of another function and a third function’s derivative.

Note: In this problem we have given the algebraic and trigonometric terms as integrands. And we used integration by parts to solve this problem. In that we have taken $u$ as $x$ and $dv$ as $\sin x$ and we have to be careful while integrating $\sin x$ because integration of $\sin x$ gives $ - \cos x$ but in differentiation, there won’t be a negative sign.