Trig functions within Trig functions

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Homework Help Overview

The problem involves evaluating the integral \(\int\frac{\sqrt{y^{2}-25}}{y} dy\) using trigonometric substitution, specifically with the substitution \(y = 5\sec(u)\). The original poster expresses confusion regarding how to handle the expression involving \(\tan(\text{arcsec}(...))\).

Discussion Character

  • Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to solve the integral using trigonometric substitution but is uncertain about how to simplify \(\tan(\text{arcsec}(y/5))\). Some participants suggest using a right-angled triangle to visualize the relationship between the trigonometric functions involved.

Discussion Status

Participants are exploring different methods to clarify the relationship between the trigonometric functions. One participant mentions previous inconsistencies with answers compared to a textbook, indicating a need for further investigation into the approach. Another participant offers a trigonometric identity to aid in the conversion of functions.

Contextual Notes

The original poster has expressed a lack of understanding regarding the handling of trigonometric functions nested within other trigonometric functions, indicating a potential gap in foundational knowledge on this topic.

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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.
 
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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.
 
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.
 
Use
[tex] 1+\tan^{2}x=\sec^{2}x[/tex]
To comvert the tan into a sec.
 

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