Chapter 7.3, Problem 8E

To determine

To find the indefinite integral.

Answer to Problem 8E

  $\text{The indefinite integral is, }{\displaystyle \int x^2\cos \left(\frac{x}{2}\right)dx}=2x^2\sin \left(\frac{x}{2}\right)+8x\cos \left(\frac{x}{2}\right)-16\sin \left(\frac{x}{2}\right)+C.$

Explanation of Solution

Given information: $\text{The equation is},{\displaystyle \int x^2{e}^{-x}dx}.$

Concept used: Tabular integration.

  $\begin{array}{l}\text{We have seen that integrals of the form }{\displaystyle \int f\left(x\right)}g\left(x\right)dx\text{ in which }f\text{ can be differentiated }\\ \text{repeatedly to become zero and }g\text{ can be integrated repeatedly without difficulty, are natural }\\ \text{candidates for integration by parts}\text{. However, if many repetitions are required, the calculation }\\ \text{can be cumbersome}\text{.}\end{array}$

Calculation : Using Tabular intergration, indefinite integral is obtained as,

  $\text{With }f\left(x\right)=x^2\text{ and }g\left(x\right)=\cos \left(\frac{x}{2}\right),\text{ we list:}$

  

  $\begin{array}{l}\text{We combine the products of the functions connected by the arrows according to the }\\ \text{operation signs above the arrows to obtain,}\\ {\displaystyle \int x^2\cos \left(\frac{x}{2}\right)dx}=x^2\left[2\sin \left(\frac{x}{2}\right)\right]-2x\left[-4\cos \left(\frac{x}{2}\right)\right]+2\left[-8\sin \left(\frac{x}{2}\right)\right]+C\\ =2x^2\sin \left(\frac{x}{2}\right)+8x\cos \left(\frac{x}{2}\right)-16\sin \left(\frac{x}{2}\right)+C\\ \text{Hence, the indefinite integral is, }{\displaystyle \int x^2\cos \left(\frac{x}{2}\right)dx}=2x^2\sin \left(\frac{x}{2}\right)+8x\cos \left(\frac{x}{2}\right)-16\sin \left(\frac{x}{2}\right)+C.\end{array}$