Int sin[sqrt[t]]

  • #1
SplinterIon
17
0
I'm drawing a blank as to how to approach this one. I've been looking over all trig identities and substitutions I could possibly make - but to no avail :cry:.

[tex]
\int_{1}^{x^2} \sin{(\sqrt{t})} \ dt
[/tex]
 

Answers and Replies

  • #2
Try making a u-substitution for the square root of t.
 
  • #3
Orion1
973
3
indefinite integral...


The closest identity that I could determine is:

identity:
[tex]\int u \sin u \; du = \sin u - u \cos u + C[/tex]

[tex]\int \sin \sqrt{t} \; dt = 2 \left( \sin \sqrt{t} - \sqrt{t} \cos \sqrt{t} \right) + C[/tex]
 
  • #4
HallsofIvy
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MalleusScientiarum said:
Try making a u-substitution for the square root of t.

Good try but it looks to me like that gives you
[tex]\int u^{\frac{1}{2}}sin u du[/tex] which doesn't look any more hopeful.

I'd be willing to bet that this doesn't have an elementary anti-derivative.
 
  • #5
dextercioby
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Don't bet too much, Halls.

[tex] \sqrt{t}=u [/tex]

implies [itex] t=u^{2} \ \mbox{and} \ dt= 2 u du [/itex]

and the antiderivative becomes

[tex] \int 2 u \sin u \ du [/tex] which can be easily tackled with the part integration method.

Daniel.
 
  • #6
HallsofIvy
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One of these days, I really have to learn algebra!
 

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