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Trig functions within Trig functions

  • Thread starter Cajstyle
  • Start date
  • #1
2
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Homework Statement



I'm given the problem: [tex]\int[/tex][tex]\frac{\sqrt{y^{2}-25}}{y}[/tex] dy


Homework Equations



I set y = 5sec(u), and solve for the subsitution.


The Attempt at a Solution



At the culmination of my solution, I achieve: 5tan(u) - 5u + C

Here is my dilemma, u stands for arcsec([tex]\frac{y}{5}[/tex]) -- How do I deal with tan(arcsec(..)), or any "trig within trig" setup? I've gone all this time without actually ever figuring out how to solve for these.
 

Answers and Replies

  • #2
954
117
You can use a right-angled triangle to aid you in visualizing what is going on. For instance, arcsec(y/5) returns an angle in a right-angled triangle where the adjacent side is of length 5 and the hypotenuse is of length y. Then, you can easily calculate the tangent of that angle.
 
  • #3
2
0
I was previously using this as a method to solve them but I ran into a few instances of having different answers than the book. I'll look further into it, thank you for the response.
 
  • #4
hunt_mat
Homework Helper
1,720
16
Use
[tex]
1+\tan^{2}x=\sec^{2}x
[/tex]
To comvert the tan into a sec.
 

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