Integrating $\int x^2\cos\left({\frac{x}{2}}\right)dx$ by parts

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Discussion Overview

The discussion revolves around the integration of the function $\int x^2\cos\left({\frac{x}{2}}\right)dx$ using integration by parts (IBP). Participants explore the steps involved in applying IBP multiple times, addressing both the polynomial and trigonometric components of the integrand.

Discussion Character

  • Mathematical reasoning
  • Homework-related
  • Technical explanation

Main Points Raised

  • One participant initiates the integration process by selecting $u={x}^{2}$ and $dv=\cos{\left(\frac{x}{2}\right)}dx$, calculating $du=2x dx$ and $v=2\sin\left({\frac{x}{2}}\right)$.
  • Another participant confirms the initial setup is correct and encourages further progress.
  • A participant presents the integration by parts formula and derives an expression involving another integral: $2x^2\sin\left({\frac{x}{2}}\right)-4\int x \sin\left({\frac{x}{2}}\right)dx$.
  • One participant notes that the original integrand consists of a polynomial and a trigonometric function, suggesting that repeated application of IBP will reduce the polynomial degree until it becomes a constant.
  • A participant attempts to integrate $\int x\sin{\left(\frac{x}{2}\right)}dx$ using IBP, defining $u=x$ and $dv=\sin\left({\frac{x}{2}}\right)dx$, leading to a derived expression involving cosine.
  • Another participant presents a final expression for the integral: $2x^2\sin\left({\frac{x}{2}}\right)+8x\cos\left({\frac{x}{2}}\right)-16\sin\left({\frac{x}{2}}\right)$, expressing hope that it is correct.
  • A subsequent post reiterates the final expression, adding a constant of integration.
  • One participant reflects on a tendency to forget certain steps in the integration process.

Areas of Agreement / Disagreement

Participants generally agree on the steps taken in the integration process, but there is no explicit consensus on the correctness of the final answer, as some participants express uncertainty.

Contextual Notes

There are indications of missing assumptions or steps in the integration process, particularly regarding the handling of the integrals after applying IBP multiple times. The discussion does not resolve these aspects.

karush
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$\int x^2\cos\left({\frac{x}{2}}\right)dx$
$u={x}^{2}\ dv=\cos{\left(\frac{x}{2}\right)}dx$
$du=2x dx\ v=\int\cos\left({\frac{x}{2}}\right)dx=2\sin\left({\frac{x}{2}}\right)$

Integrat by parts, just seeing if getting started ok
 
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karush said:
$\int x^2\cos\left({\frac{x}{2}}\right)dx$
$u={x}^{2}\ dv=\cos{\left(\frac{x}{2}\right)}dx$
$du=2x dx\ v=\int\cos\left({\frac{x}{2}}\right)dx=2\sin\left({\frac{x}{2}}\right)$

Integrat by parts, just seeing if getting started ok

That is correct! Go on ...
 
$$uv-\int v \ du$$

So

$$2x^2\sin\left({\frac{x}{2}}\right)-4\int x \sin\left({\frac{x}{2}}\right)dx$$

Look like another round of parts..
 
Yup, you should notice that your original integrand consists of two functions, a polynomial and a trigonometric function. Each time you apply IBP, the degree of your polynomial should decrement by 1, eventually turning into a constant. Then, you will encounter either $\sin\left({\frac{x}{2}}\right)$ or $\cos\left({\frac{x}{2}}\right)$ which you can easily integrate.
 
$\int x\sin{\left(\frac{x}{2}\right)}du$
$u=x\ dv=\sin\left({\frac{x}{2}}\right)dx$
$du=dx\ v=\int\sin{\left(\frac{x}{2}\right)}dx=-2\cos{\left(\frac{x}{2}\right)}$
$4\left[-2x\cos\left({\frac{x}{2}}\right)-4\sin\left({\frac{x}{2}}\right)\right]$
 
$2x^2\sin\left({\frac{x}{2}}\right)+8x\cos\left({\frac{x}{2}}\right)-16\sin\left({\frac{x}{2}}\right)$
My final answer (I hope)
 
karush said:
$2x^2\sin\left({\frac{x}{2}}\right)+8x\cos\left({\frac{x}{2}}\right)-16\sin\left({\frac{x}{2}}\right)$
My final answer (I hope)

+ C
 
I seem to forget that to often...
 

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