Evaluating Square Root Integrals (Trig. Substitution?)

[mill4864]

Instructions: Evaluate the Integral.

[ ( 1 ) / ( sqroot ( t^2 - 6t + 13 ) ] dt




I apologize for not being able to write this in mathematical form. I typed up the integral using Microsoft Word >> Microsoft Equation 3.0, but when I tried copying and pasting the equation into my post, it didn't work.




The problem is in a section of the book called "Trigonometric Substitution", so I'm assuming that we have to use trigonometric substitution to solve the integral. So what I did was:

(1) Turned ( t^2 - 6t + 13 ) into [ ( t - 3 )^2 + 4 ]. I think that is "completing the square".

(2) Substituted u = ( t - 3 ), just to make the integral simpler, I guess.

[ ( 1 ) / ( sqroot ( u^2 + 4 ) ] du

(3) Substituted u = ( 2 tanx ) and du = [ ( 2 (secx)^2 ) dx ]

[ ( 2 (secx)^2 ) / ( sqroot ( 4 ((tanx)^2 + 1))) ] du

(4) Used the trigonometric identity (tanx)^2 + 1 = (secx)^2

[ ( 2 (secx)^2 ) / ( sqroot ( 4 (secx)^2 ) ] du

(5) Simplified further to get the integral of (secx).

I don't think we are supposed to use the integration formula that the integral of (secx)dx = ln |secx + tanx| + c. And I don't understand how to plug the t's back in for the x's because I think we'd have to use inverse trigonometric functions and I don't think that's what I'm suppsed to be getting, but I could be wrong.

 

[BigGlenntheHeavy]

$\displaystyle \int \frac{dt}{[t^{2}-6t+13]^{1/2}} \ = \ \int \frac{dt}{[(t-3)^{2}+4]^{1/2}} \ Let \ u \ = \ t-3, \ then \ du \ = \ dt.$

$\displaystyle = \ \int \frac{du}{[u^{2}+4]^{1/2}}, \ Now \ let \ u \ = \ 2tan(\theta), \ then \ du \ = \ 2sec^{2}(\theta)d\theta.$

$\displaystyle Ergo, \ we \ have \ \int \frac{2sec^{2}(\theta)d\theta}{[4tan^{2}(\theta)+4]^{1/2}}$

$\displaystyle Which \ simplifies \ to \ \int \frac{sec^{2}(\theta)d\theta}{sec(\theta)} \ = \ \int sec(\theta)d\theta$

$\displaystyle \int sec(\theta)d\theta \ = \ ln|sec(\theta)+tan(\theta)|+C_1.$

$\displaystyle Resubstituting \ (tan(\theta)=\frac{u}{2}), \ gives \ ln\bigg|\frac{[u^{2}+4]^{1/2}}{2}+\frac{u}{2}\bigg|+C_1$

$\displaystyle Resubstituting \ again \ (u=t-3) \ gives: \ ln\bigg|\frac{[t^2-6t+13]^{1/2}+t-3}{2}\bigg|+C_1$

\(\displaystyle = \ ln|[t^{2}-6t+13]^{1/2}}+t-3|-ln(2)+C_1 \ = \ln|[t^{2}-6t+13]^{1/2}+t-3|+C, \ C \ = \ -ln(2)+C_1.\)

$\displaystyle Therefore, \ after \ all \ is \ said \ and \ done, \ we \ have \ \int \frac{dt}{[t^{2}-6t+13]^{1/2}} \ = \ln|[t^{2}-6t+13]^{1/2}+t-3|+C$

$\displaystyle Check: \ D_t \ [ln|(t^{2}-6t+13)^{1/2}+t-3|+C] \ = \ \frac{(1/2)(t^{2}-6t+13)^{-1/2}(2t-6)+1}{(t^{2}-6t+13)^{1/2}+t-3}$

$\displaystyle = \ \frac{(t^{2}-6t+13)^{-1/2}(t-3)+1}{(t^{2}-6t+13)^{1/2}+t-3}$

$\displaystyle = \ \frac{\frac{t-3}{(t^{2}-6t+13)^{1/2}}+1}{(t^{2}-6t+13)^{1/2}+t-3}$

\(\displaystyle = \ \frac{\frac{t-3+(t^{2}-6t+13)^{1/2}}{(t^{2}-6t+13)^{1/2}}}{\frac{(t^{2}-6t+13)^{1/2}+t-3}{1}}}\)

$\displaystyle = \ \frac{t-3+(t^{2}-6t+13)^{1/2}}{(t^{2}-6t+13)^{1/2}} \ * \ \frac{1}{(t^{2}-6t+13)^{1/2}+t-3} \ = \ \frac{1}{(t^{2}-6t+13)^{1/2}}, \ QED$

 

[Deleted member 4993]

Glen,

That was some display of your skills at "tex" - not to diminish your skills at solving the problem...

 

[mmm4444bot]

… I typed up the integral using Microsoft Word … but when I tried copying and pasting the equation into my post, it didn't work …

We can upload image files, to these boards. (I'm thinking that the file format needs to be either .JPG or .GIF)

Here's how to do it, using the Microsoft Windoze operating system:

(1) Display on your monitor the relevant part of whatever you want to upload, at the size you want, and then hit Alt-PrtSc. (This copies the active window contents to the clipboard.)

(2) Open MSPaint, and paste the screen-shot into a new document.

(3) Resize and/or crop the MSPaint document, to show only the relevant part of the image, and then save the file in jpeg format.

(4) Upload attachment here.

(5) Place the cursor within the text of your post ? right where you want the image to appear ? and then click the [Place inline] button.

Let me know, if you need more help with this process.