How to integrate this integral?

In summary, the integration \int_{0}^{\sqrt{2}h}\frac{x}{\sqrt{h^2+x^2-\sqrt{2}hx}}dx can be solved by first completing the square in the denominator and then using the substitution u = x - \frac{\sqrt{2}}{2}h. The integral can then be broken up into two parts, which can be solved using the substitutions t = \sqrt{u^2 + \frac{h^2}{2}} and u = \frac{h\tan v}{\sqrt{2}}, respectively. The final solution is \sqrt{x^2-\sqrt{2}hx+h^2}+\frac{\sqrt{2
  • #1

Homework Statement


Perform the following integration:

[tex]\int_{0}^{\sqrt{2}h}\frac{x}{\sqrt{h^2+x^2-\sqrt{2}hx}}dx[/tex]

Homework Equations



The solution (via wolframalpha.com) is:

[tex][\sqrt{h^2+x^2-\sqrt{2}hx}+\frac{h}{\sqrt{2}}ln(2(\sqrt{h^2+x^2-\sqrt{2}hx}+x)-\sqrt{2}h)[/tex]

Evaluated from 0 to sqrt(2)h

Here is the link: http://www.wolframalpha.com/input/?i=x/[(h^2+x^2-(2^(1/2))*x*h)^(1/2)]

The Attempt at a Solution



I immediately looked at a substitution and an integration by parts. They left me worse off than before. Then, I went to wolframalpha in order to see how to do it. They could not show any steps. I am not sure how to integrate this by hand. Any advice?

If this is an integral that you simply look up in a table, could someone point me to the methods used (no matter how complex) that will solve this?

Thanks for reading.
 
Physics news on Phys.org
  • #2
First complete the square in the denominator:
[tex]x^2- \sqrt{2}hx+ h^2= x^2- 2\sqrt{2}hx+ \frac{h^2}{2}- \frac{h^2}{2}+ h^2[/tex]
[tex](x- \frac{\sqrt{2}}{2}h)^2+ \frac{h^2}{2}[/tex]

Make the substitution
[tex]u= x- \frac{\sqrt{2}}{2}h[/tex]
 
  • #3
CalcYouLater said:

Homework Statement


Perform the following integration:

[tex]\int_{0}^{\sqrt{2}h}\frac{x}{\sqrt{h^2+x^2-\sqrt{2}hx}}dx[/tex]

Homework Equations



The solution (via wolframalpha.com) is:

[tex][\sqrt{h^2+x^2-\sqrt{2}hx}+\frac{h}{\sqrt{2}}ln(2(\sqrt{h^2+x^2-\sqrt{2}hx}+x)-\sqrt{2}h)[/tex]

Evaluated from 0 to sqrt(2)h

Here is the link: http://www.wolframalpha.com/input/?i=x/[(h^2+x^2-(2^(1/2))*x*h)^(1/2)]

The Attempt at a Solution



I immediately looked at a substitution and an integration by parts. They left me worse off than before. Then, I went to wolframalpha in order to see how to do it. They could not show any steps. I am not sure how to integrate this by hand. Any advice?

If this is an integral that you simply look up in a table, could someone point me to the methods used (no matter how complex) that will solve this?

Thanks for reading.
Try completing the square under the radical. Then do a substitution.

[tex]\sqrt{h^2+x^2-\sqrt{2}hx}}\ =\ \sqrt{x^2-2\frac{h}{\sqrt{2}}x+\frac{h^2}{2}-\frac{h^2}{2}+h^2}\,}[/tex]

[tex]=\sqrt{\left(x-\frac{h}{\sqrt{2}}\right)^2+\frac{h^2}{2}\,}[/tex]
 
  • #4
Thank you, HallsofIvy and SammyS.

Following HallsofIvy's suggestion for the value of u, I did the following:

[tex]\int_{0}^{\sqrt{2}h}\frac{x}{\sqrt{h^2+x^2-\sqrt{2}hx}}dx\rightarrow{\int\frac{u+\frac{\sqrt{2}}{2}h}{\sqrt{(u^{2}+\frac{h^{2}}{2})}}du\rightarrow{\int\frac{u+\frac{\sqrt{2}}{2}h}{\sqrt{(u-\frac{\sqrt{2}}{2}h)(u+\frac{\sqrt{2}}{2}h)}}du[/tex]




Then:

[tex]{\int\frac{u+\frac{\sqrt{2}}{2}h}{\sqrt{(u-\frac{\sqrt{2}}{2}h)(u+\frac{\sqrt{2}}{2}h)}}du={\int\sqrt{\frac{u+\frac{\sqrt{2}}{2}h}{u-\frac{\sqrt{2}}{2}h}}du[/tex]

Here is where things began to get circular. I tried integration by parts two times in succession, this lead me back to the original integral (the last one listed above). It seems I either over-complicated the situation, or missed a key step.
 
Last edited:
  • #5
CalcYouLater said:
Thank you, HallsofIvy and SammyS.

Following HallsofIvy's suggestion for the value of u, I did the following:

[tex]\int_{0}^{\sqrt{2}h}\frac{x}{\sqrt{h^2+x^2-\sqrt{2}hx}}dx\rightarrow{\int\frac{u+\frac{\sqrt{2}}{2}h}{\sqrt{(u^{2}+\frac{h^{2}}{2})}}du\rightarrow{\int\frac{u+\frac{\sqrt{2}}{2}h}{\sqrt{(u-\frac{\sqrt{2}}{2}h)(u+\frac{\sqrt{2}}{2}h)}}du[/tex]

[tex]u^{2}+\frac{h^{2}}{2}\neq\left(u-\frac{\sqrt{2}}{2}h\right)\left(u+\frac{\sqrt{2}}{2}h\right)[/tex]


[tex]\left(a+b\right)\left(a-b\right)=a^2-b^2\neq a^2+b^2[/tex]

Break the integrand up:

[tex]\int\frac{u+\frac{\sqrt{2}}{2}h}{\sqrt{u^{2}+\frac{h^{2}}{2}}}\ du=\int\frac{u}{\sqrt{u^{2}+\frac{h^{2}}{2}}}\ du\ +\ \left(\frac{\sqrt{2}}{2}h\right)\int\frac{1}{\sqrt{u^{2}+\left(\frac{h}{\sqrt{2}}\right)^2}}\ du[/tex]

The first can be done by the substitution [tex]t= u^2+\frac{h^2}{2}\,,[/tex] or even better, [tex]t= \sqrt{u^2+\frac{h^2}{2}}\,.[/tex]

The second by the substitution [tex]u= \frac{h\,\tan(v)}{\sqrt{2}}\,.[/tex]
 
Last edited:
  • #6
Thanks SammyS! After a bit I was able to work out the answer as:

[tex]\sqrt{x^2-\sqrt{2}hx+h^2}+\frac{\sqrt{2}h}{2}[ln((\frac{\sqrt{2}h}{2})(\sqrt{x^2-\sqrt{2}hx+h^2}+x-\frac{\sqrt{2}h}{2}))][/tex]

Which after evaluating from 0 to sqrt(2)h gives me:

[tex]\frac{\sqrt{2}h}{2}[ln(1+\sqrt{2})-ln(\sqrt(2)-1)][/tex]

Which is the same result as listed in the initial problem. Thanks again! It was good to sharpen my integration skills.
 

What is an integral?

An integral is a mathematical concept that represents the accumulation or total of a given function over a certain interval. It is the inverse operation of differentiation and is commonly used in calculus and other areas of mathematics.

Why do we need to integrate?

Integrals are used to find the area under a curve, which has many real-world applications such as calculating volume, distance, and work. They also allow us to solve differential equations, which are used to model many physical phenomena.

How do I know which integration method to use?

The method for integrating a given integral depends on the form of the function. Some common methods include substitution, integration by parts, and trigonometric substitutions. It is important to analyze the integral and try different methods to find the most efficient way to solve it.

What are the limits of integration?

The limits of integration are the lower and upper bounds of the integral, which represent the starting and ending points of the interval over which the function is being integrated. These limits are typically denoted as a and b in the notation ∫ab f(x) dx.

Can I use a calculator to integrate?

While some calculators may have integration capabilities, it is important to understand the concepts and methods of integration rather than relying solely on technology. Additionally, certain integrals may be too complex for a calculator to solve accurately.

Suggested for: How to integrate this integral?

Replies
25
Views
1K
Replies
5
Views
906
Replies
9
Views
914
Replies
2
Views
730
Replies
44
Views
3K
Replies
16
Views
981
Replies
2
Views
836
Back
Top