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Trigonometric antiderivatives

  • Thread starter bobsmith76
  • Start date
  • #1
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Homework Statement



Screenshot2012-02-08at54843AM.png


I don't see how they're going from

sec x cot x to csc x

and

csc x tan x to sec x

The derivative of sec x is sec x tan x not csc x tan x
and the derivative of csc x is -csc x cot x

and if it's an identity then they didn't care to inform me of that at this website

http://www.sosmath.com/trig/Trig5/trig5/trig5.html
 

Answers and Replies

  • #2
Curious3141
Homework Helper
2,843
86

Homework Statement



Screenshot2012-02-08at54843AM.png


I don't see how they're going from

sec x cot x to csc x

and

csc x tan x to sec x
Express sec x, csc x, cot x and tan x in terms of sin x and cos x first.

The derivative of sec x is sec x tan x not csc x tan x
and the derivative of csc x is -csc x cot x
Sure, but what's the relevance of this? They weren't integrating at that stage, merely simplifying using trig identities.
 
  • #3
eumyang
Homework Helper
1,347
10

Homework Statement



Screenshot2012-02-08at54843AM.png


I don't see how they're going from

sec x cot x to csc x

and

csc x tan x to sec x

The derivative of sec x is sec x tan x not csc x tan x
and the derivative of csc x is -csc x cot x

and if it's an identity then they didn't care to inform me of that at this website

http://www.sosmath.com/trig/Trig5/trig5/trig5.html
You should not only recognize the trig identities, but different variations of them. For instance,
[itex]\tan \theta = \frac{\sin \theta}{\cos \theta}[/itex].
But it is also true that
[itex]\tan \theta = \sin \theta \sec \theta[/itex].
 

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