How do i figure out this integral

  • #1
afcwestwarrior
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Homework Statement


∫ 1/cos theta
before it was ∫ sec theta but i know that sec theta is equal to 1/cos theta

would i write it as 1/2 ∫ cos theta

and then 1/2 sin theta
 

Answers and Replies

  • #2
rock.freak667
Homework Helper
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try multiplying sec(theta) by

[tex]\frac{sec\theta +tan\theta}{sec\theta +tan\theta}[/tex]


then try a substitution.
 
  • #3
afcwestwarrior
457
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where did you get sec theta + tan theta/ sec theta+ tan theta from
 
  • #4
afcwestwarrior
457
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is this what i do (sec theta) (sec theta+ tan theta)/ Sec theta( sec theta + tan theta)
 
  • #5
rock.freak667
Homework Helper
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is this what i do (sec theta) (sec theta+ tan theta)/ Sec theta( sec theta + tan theta)


You should get

[tex]\frac{sec\theta(sec\theta +tan\theta)}{sec\theta + tan\theta}[/tex]


The expand out the numerator.


It's an easier way to find the integral faster rather than another method.
 
  • #6
afcwestwarrior
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Ok thank you.
 
  • #7
HallsofIvy
Science Advisor
Homework Helper
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By the way, [itex]\int dx/cos(x)[/itex] involves an odd power of cosine so the "standard" method for that situation will work. Multiply both numerator and denominator by cos(x) to get
[tex]\int \frac{dx}{cos x}= \int \frac{cos x dx}{cos^2 x}= int \frac{cos x dx}{1- sin^2 x}[/tex]

Now let u= sin x so du= cos x dx and 1- sin2 x= 1- u2

[tex]\int \frac{dx}{cos x}= \int \frac{du}{1- u^2}= \int \frac{du}{(1- u)(1+ u)}[/itex]

and you can use "partial fractions" to integrate that.
 

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