Use trigonometric substitution to verify

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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.
:)
 
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