Thanks! Glad you found it helpful.

  • Context: MHB 
  • Thread starter Thread starter karush
  • Start date Start date
  • Tags Tags
    Dx
Click For Summary

Discussion Overview

The discussion revolves around the evaluation of the integral $$\int \frac{{x}^{3 }}{\sqrt{4x ^2 - 1}} \ dx$$ using various methods, including substitution and integration by parts (IBP). Participants explore different approaches and share their calculations, focusing on the steps involved in simplifying and solving the integral.

Discussion Character

  • Technical explanation, Mathematical reasoning, Debate/contested

Main Points Raised

  • One participant presents a solution using substitution, leading to the expression $$\frac{1}{32}\displaystyle \int\dfrac{u+1}{\sqrt{u}}\,\mathrm{d}u$$.
  • Another participant corrects a notation regarding the substitution, stating that $$x = \left( \frac{u + 1}{4} \right)^{\frac{1}{2}}$$ is the correct form, but acknowledges that the integrand remains correct.
  • Subsequent posts reiterate the integration steps, with one participant expressing uncertainty about copying notes accurately, yet arriving at the same result for the integral.
  • A different approach using integration by parts is introduced, where a participant defines $$u=x^2$$ and derives an expression for $$dv$$, leading to a new formulation of the integral.
  • Questions arise regarding the origin of $$dv$$ in the integration by parts method, with multiple participants seeking clarification.
  • A later reply expresses appreciation for the IBP solution, indicating a positive reception of that method.

Areas of Agreement / Disagreement

Participants do not reach a consensus on a single method for solving the integral, as multiple approaches are presented and discussed. There is also some confusion regarding the notation and steps in the integration by parts method.

Contextual Notes

Some participants express uncertainty about specific steps in their calculations, particularly in relation to notation and the derivation of $$dv$$. The discussion includes various methods without resolving which is superior or more accurate.

karush
Gold Member
MHB
Messages
3,240
Reaction score
5
Whitman 8.3.12
$$\int \frac{{x}^{3 }}{\sqrt{4x ^2 - 1}} \ dx =
\frac{\left(2{x}^{2}+1\right)\sqrt{4{x}^{2}-1}}{24}$$
$$u=4x^2 - 1 \ \ \ \ du=8x \ dx \ \ \ x=\left(\frac{u-1}{4 }\right)^\frac{1}{2}$$
Substitute and simplify
$$\frac{1}{32}\displaystyle \int\dfrac{u+1}{\sqrt{u}}\,\mathrm{d}u$$
 
Last edited:
Physics news on Phys.org
Re: Whitman 8.3.12 \int \frac{{x}^{3 }}{\sqrt{4x ^2 - 1}} \ dx =

karush said:
Whitman 8.3.12
$$\int \frac{{x}^{3 }}{\sqrt{4x ^2 - 1}} \ dx =
\frac{\left(2{x}^{2}+1\right)\sqrt{4{x}^{2}-1}}{24}$$
$$u=4x^2 - 1 \ \ \ \ du=8x \ dx \ \ \ x=\left(\frac{u-1}{4 }\right)^\frac{1}{2}$$
Substitute and simplify
$$\frac{1}{32}\displaystyle \int\dfrac{u+1}{\sqrt{u}}\,\mathrm{d}u$$

Actually it's $\displaystyle \begin{align*} x = \left( \frac{u + 1}{4} \right) ^{\frac{1}{2}} \end{align*}$, but it appears that you had this right anyway as you have gotten the correct integrand.

So now write $\displaystyle \begin{align*} \frac{1}{32} \int{ \frac{u + 1}{\sqrt{u}}\,\mathrm{d}u } = \frac{1}{32}\int{ \left( u^{\frac{1}{2}} + u^{-\frac{1}{2}} \right) \,\mathrm{d}u } \end{align*}$ and integrate.
 
Yeah didn't copy my notes to good soooo..
$$\frac{1} {32}\left[\frac{2{u}^{3/2}}{3 }+2u^{1/2}\right]
=\left[\frac{\sqrt{u}\left(u+3\right)}{48}\right]$$
Back substittute $u=4{x}^{2}-$1 then

$I =
\frac{\left(2{x}^{2}+1\right)\sqrt{4{x}^{2}-1}}{24}$

🍸🍸🍸🍸🍸🍸🍸🍸🍸🍸🍸🍸🍸🍸🍸🍸🍸🍸
 
karush said:
yeah didn't copy my notes to good soooo..
$$\frac{1} {32}\left[\frac{2{u}^{3/2}}{3 }+2u^{1/2}\right]
=\left[\frac{\sqrt{u}\left(u+3\right)}{48}\right]$$
back substittute $u=4{x}^{2}-$1 then

$i =
\frac{\left(2{x}^{2}+1\right)\sqrt{4{x}^{2}-1}}{24}$

🍸🍸🍸🍸🍸🍸🍸🍸🍸🍸🍸🍸🍸🍸🍸🍸🍸🍸

+c :p
 
Let's try IBP...we are given:

$$I=\int \frac{x^3}{\sqrt{4x^2-1}},dx$$

Let:

$$u=x^2\implies du=2x\,dx$$

$$dv=\frac{1}{8}\cdot\frac{8x}{\sqrt{4x^2-1}}\,dv\implies v=\frac{1}{4}\sqrt{4x^2-1}$$

Hence:

$$I=\frac{x^2}{4}\sqrt{4x^2-1}-\frac{1}{16}\int 8x\sqrt{4x^2-1}\,dx$$

$$I=\frac{x^2}{4}\sqrt{4x^2-1}-\frac{1}{24}\left(4x^2-1\right)^{\frac{3}{2}}+C$$

$$I=\frac{\sqrt{4x^2-1}}{24}\left(6x^2-\left(4x^2-1\right)\right)+C$$

$$I=\frac{\sqrt{4x^2-1}}{24}\left(2x^2+1\right)+C$$
 
Where does $dv \ \ \ \ \ $ come from?
 
karush said:
Where does $dv \ \ \ \ \ $ come from?

The original integral may be written as:

$$I=\int u\,dv$$
 
Got it

Liked the way that was solved😃
 

Similar threads

  • · Replies 4 ·
Replies
4
Views
3K
  • · Replies 6 ·
Replies
6
Views
3K
  • · Replies 6 ·
Replies
6
Views
3K
  • · Replies 4 ·
Replies
4
Views
4K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 8 ·
Replies
8
Views
2K
  • · Replies 2 ·
Replies
2
Views
3K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 12 ·
Replies
12
Views
3K