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Integration by substitution find the area under the graph

  • Thread starter bobred
  • Start date
  • #1
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Homework Statement



I have the function below which i need to find the area under the graph.

Homework Equations



[tex]\int_{ - \frac {\pi}{4}}^{\frac {\pi}{3}}\frac {2\sec x}{2 + \tan x}dx[/tex]

The Attempt at a Solution



I can simplify it to

[tex]\int_{ - \frac {\pi}{4}}^{\frac {\pi}{3}}\frac {2}{2\cos x + \sin x}[/tex]

But I am at a loss as to where to go from here, I've tried all sorts of things. I have Mathcad which has given me the answer, but I want to be able to do it myself. I have been shown the Weierstrass substitution method which I can follow, but it is something that isn't covered by the course.

Thanks
 
Last edited:

Answers and Replies

  • #2
Dick
Science Advisor
Homework Helper
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Now do the substitution. t=tan(x/2). I.e. sin(x)=2t/(1+t^2) etc. Try to integrate the resulting rational function. What's stopping you from continuing? It's sounds like you know the right method.
 
  • #3
173
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Hi

Yeah, I get to

2/t^2-t-1

and then I'm not sure what to do.
 
  • #4
Dick
Science Advisor
Homework Helper
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I seem to be getting -2/(t^2-t-1). But now you complete the square. Write the quadratic as (t-a)^2-b^2. What next?
 
  • #5
173
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Hi

I decided to try another route using trig identities.

[tex]\frac{2\sec x}{2+\tan x}=\frac{2}{\sqrt{5}\cos(\theta-x)}=\frac{2}{\sqrt{5}}\cdot\frac{1}{\cos(\theta-x)}=\frac{2}{\sqrt{5}}\cdot\sec(\theta-x)[/tex]

where [tex]\theta=\arctan\left(1/2\right)[/tex]

[tex]\int\frac{2\sec x}{2+\tan x}=\frac{2}{\sqrt{5}}\ln\left|\sec (-\arctan\left(1/2\right)+x)+\tan (-\arctan\left(1/2\right)+x)\right|+c[/tex]
 

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