Square root integration problem

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SUMMARY

The forum discussion centers on the integration of the function f(t) = t³/√(t²+1). The initial approach involved substituting u = √(t²+1), leading to the transformation of the integral into a simpler form. A participant suggested an alternative substitution, u = t² + 1, which simplifies the integration process further. Both methods ultimately yield the same result, confirming the validity of the original approach.

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kukumaluboy
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Homework Statement



Integrate : f(t) = t3/ sqrt(t2+1)

Homework Equations


The Attempt at a Solution


t2* t / sqrt(t2+1)let u = sqrt(t2+1)
u=(t2+1)0.5
u2 = t2 +1
t2 = u2 -1 -Eq 1u=(t2-1)0.5
du = t/sqrt(t2-1) dt
dt = sqrt(t2-1)/t du -Eq 2

Hence Substituting 1 and 2
Integrate : f(t)
= Integrate (u^2 -1) du

Is this way correct? If not can give me the right way
 
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Looks good to me. I think I would have been inclined to use the simpler [itex]u= t^2+ 1[/itex], du= 2t dt so that t dt= (1/2)du and [itex]t^2= u- 1[/itex].

That would give
[tex]\int t^2\sqrt{t^2+ 1}dt= \int(u- 1)(\sqrt{u}(1/2)du= (1/2)\int u^{3/2}- u^{1/2}du[/tex]
but that will give the same thing as your integral.
 
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Alrite Thanks!
 

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