Use trigonometric substitution to verify

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SUMMARY

The discussion focuses on using trigonometric substitution to verify the integral \(\int \sqrt{a^2 - t^2} dt\) from 0 to \(\pi\). The substitution \(x = a \sin(\theta)\) is employed, leading to the transformation of the integral into \(-\int a^2 \cos^2(\theta) d\theta\). Participants highlight the importance of applying the double angle formula to simplify \(\cos^2(\theta)\) to \(\frac{1 + \cos(2\theta)}{2}\) for further evaluation. The conversation emphasizes the correct identification of constants and the manipulation of trigonometric identities to facilitate the integration process.

PREREQUISITES
  • Understanding of trigonometric identities, specifically \(\sin\) and \(\cos\) functions.
  • Familiarity with integral calculus, particularly definite integrals.
  • Knowledge of trigonometric substitution techniques in calculus.
  • Ability to apply double angle formulas in trigonometric functions.
NEXT STEPS
  • Study the application of trigonometric substitution in integrals involving square roots.
  • Learn about the double angle formulas for sine and cosine functions.
  • Explore the derivation and application of integral calculus techniques for definite integrals.
  • Practice solving integrals using various substitution methods, including trigonometric and algebraic substitutions.
USEFUL FOR

Students and educators in calculus, particularly those focusing on integral calculus and trigonometric identities, as well as anyone looking to enhance their problem-solving skills in mathematical analysis.

celeramo
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Homework Statement


Using trigonometric substitution, verify that
[tex]\int[/tex] [tex]\sqrt{a^2[-t^2}[/tex]dt (INTEGRAL FROM 0 TO [tex]\pi[/tex])=(1/2)a^2sin^{-1}(x/a)+(1/2)x[tex]\sqrt{a^2-x^2[/tex]

Sorry it doesn't seem to want to allow me to place superscript inside a square root. but inside the first sq. root is a2-t2 and the second sq. root contains a2-x2



Homework Equations



[tex]\sqrt{a^2-x^2}[/tex]

x=asin([tex]\theta[/tex])

-([tex]\pi[/tex]/2) [tex]\leq[/tex] [tex]\theta[/tex] [tex]\leq[/tex] ([tex]\pi[/tex]/2)

The Attempt at a Solution



replacing x with asin[tex]\theta[/tex] in the original integral
[tex]\int[/tex][tex]\sqrt{a^2-(asin^2\theta}[/tex]dt
multiply and factor out. Change 1-sin^2[tex]\theta[/tex] to cos^2[tex]\theta[/tex]
replace dt with dt=(-a)cos[tex]\theta[/tex] d[tex]\theta[/tex] because t=asin[tex]\theta[/tex]
End up with
[tex]\int[/tex] (a)(cos[tex]\theta[/tex])(-a)(cos[tex]\theta[/tex])d[tex]\theta[/tex]
Combine and get
-[tex]\int[/tex] a2cos2[tex]\theta[/tex] d[tex]\theta[/tex]

At this point I considered pulling the a2 out in front since, unless I'm confused, it represents a constant and changing cos2[tex]\theta[/tex] to 1-sin2[tex]\theta[/tex] but that hasn't seemed to get me any close. Any help would be much appreciated or if I've made a mistake already identifying that for me would be excellent. Please and thank you all.
:)
 
Last edited by a moderator:
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No, changing cos^2(theta) to sin^2(theta) doesn't help much. But using a double angle formula does. Change it to (1+cos(2*theta))/2
 
Thanks a lot.
 

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