[itex]sec^{2}(x)tan(x)dx[/itex] Let U = sec(x) or tan(x) ?

  • Thread starter Lebombo
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  • #1
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The integral of [itex]sec^{2}(x)tan(x)dx[/itex] is what I'm asking about.

Homework Statement





[itex]sec^{2}(x)tan(x)dx[/itex]

I can let u = tanx
then du = [itex]sec^{2}(x)[/itex]

[itex]\frac{1}{2}\int udu[/itex]

[itex]\frac{1}{2}tan^{2}(x)[/itex]


OR

I can let u = secx
then du = secxtanx dx

[itex]\int udu[/itex]

[itex]\frac{1}{2}u^{2}[/itex]

[itex]\frac{1}{2}sec^{2}(x)[/itex]


Why do both work? Which one is correct? Or are both correct?
 

Answers and Replies

  • #2
Dick
Science Advisor
Homework Helper
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The integral of [itex]sec^{2}(x)tan(x)dx[/itex] is what I'm asking about.

Homework Statement





[itex]sec^{2}(x)tan(x)dx[/itex]

I can let u = tanx
then du = [itex]sec^{2}(x)[/itex]

[itex]\frac{1}{2}\int udu[/itex]

[itex]\frac{1}{2}tan^{2}(x)[/itex]


OR

I can let u = secx
then du = secxtanx dx

[itex]\int udu[/itex]

[itex]\frac{1}{2}u^{2}[/itex]

[itex]\frac{1}{2}sec^{2}(x)[/itex]


Why do both work? Which one is correct? Or are both correct?
They both work. They differ by a constant. sec(x)^2-tan(x)^2=1. You should put a '+C' in when you write the solution to an indefinite integral. That's where the difference is.
 
  • #3
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thanks, appreciate the feedback.
 

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