Can Trig Substitution with Cosine be Used Instead of Sine?

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SUMMARY

Trig substitution can be performed using either sine or cosine, as demonstrated in the discussion regarding the integral of f(x) = √(9 - x²) / x². While textbooks predominantly favor sine substitution, cosine substitution yields valid results as well, with the only difference being the presence of a constant term that can be absorbed into the integration constant C. The choice between sine and cosine depends on convenience, with sine often being slightly easier due to its derivative not introducing negative signs.

PREREQUISITES
  • Understanding of trigonometric identities and relationships
  • Familiarity with integration techniques, specifically trigonometric substitution
  • Knowledge of the function f(x) = √(9 - x²) / x²
  • Basic calculus concepts, including integration constants
NEXT STEPS
  • Explore the derivation and application of trigonometric identities in integration
  • Learn about the implications of choosing different substitution methods in integrals
  • Study the properties of inverse trigonometric functions, particularly sin⁻¹ and cos⁻¹
  • Investigate the impact of integration bounds on the results of trigonometric substitutions
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Students and educators in calculus, mathematicians exploring integration techniques, and anyone interested in the nuances of trigonometric substitutions in calculus.

UMath1
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I was wondering if you could do a trig substitution with cosine instead of sine. All the textbooks I have referred to use a sine substitution and leave no mention as to why cosine substitution was not used. It seemed that it should work just the same, until I tried it for the following Fint [sqrt(9-x^2)]/ [x^2]. I checked to see if my answer differed by only a constant but that was not the case. I have attached pictures of my work. Can anyone tell me why it does not work?
20160130_164904.jpg
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If you use sine instead, you will end up with ##\sin^{-1}## in place of ##\cos^{-1}##. But the two expressions are related by ##\cos^{-1}x = \pi/2-\sin^{-1}x##.
 
I know but why are the answers different? Is one less valid than the other?

Btw the textbook which uses sine has the same answer but with -sin^-1(x/3) instead of cos^-1(x/3) like I have it.
 
UMath1 said:
Btw the textbook which uses sine has the same answer but with -sin^-1(x/3) instead of cos^-1(x/3) like I have it.
That's exactly the point I addressed in my previous post. Replace -sin^-1(x/3) with the equation I wrote before. You will indeed have an additional ##\pi/2## but it's a constant and hence can be absorbed into the integration constant ##C##.
 
There's really nothing magic about using sin or cos. It just depends on what is more convenient for each case. As for signs, using the relevant relations from trigonometry - like the one that blue_leaf77 mentions, you can substitute sin for cos and vice versa and find the appropriate sign.
 
The function being integrated is

f(x) = √(9 - x2) / x2

. This is defined for 0 < |x| ≤ 3.

When making a substitution we want to choose an interval where f(x) makes sense, and the easiest one is 0 < x ≤ 3.

We also want to choose a substitution that takes the same values that f(x) does over the interval of definition, and that's between 0 and 3.

Each of y = 3 sin(x) and y = 3 cos(x) satisfy this condition, so either one can be used for the substitution.

Using 3 sin(x) to substitute might be a tiny bit easier than cosine because its derivative is 3 cos(x), and this does not introduce negative signs.
 
Ok...I see it now. I tried some test bounds of integration and got the same answer from both options.
 

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