Integrate that by parts for integral

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SUMMARY

The integral \(\int 9 \sin(\sqrt{t+1}) dt\) can be solved using integration by parts after a substitution. First, substitute \(u = \sqrt{t+1}\), leading to \(du = \frac{1}{2}(t+1)^{-1/2} dt\) or \(dt = 2u du\). This transforms the integral into \(9\int u \sin(u) du\), which is then integrated by parts to find the solution.

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i need help taking this integral

[tex]\int[/tex] 9sin[tex]\sqrt{t+1}[/tex]


help please! i don't even know where to start :[
 
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First write it correctly:
[tex]\int 9 sin(\sqrt{t+1}) dt[/tex]

Make the obvious substitution [itex]u= \sqrt{t+1}= (t+1)^{1/2}[/itex] so that [itex]du= (1/2)(t+1)^{-1/2}dt[/itex] or 2udu= dt.
(Equivalently, u2= t+ 1.)

That changes your integral to
[tex]9\int u sin(u)du[/tex]
and you can integrate that by parts.
 
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