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Let 
. Show that (2n-1)In + 2(n-2)In-3 = 2sqrt(2).
Then compute I8.
Source: nrich.maths.org
We'll integrate using the "integration by parts" method,
The trick in this method is always to pick the right way to split up the
integral into u and dv. Here, we decided to go with 
just because it's a natural to integrate using the substitution w=x^3+1, and
that leaves
.
Now, integrating dv, and then using the identity sqrt(u)=u/sqrt(u) to "rephrase"
the v with a common denominator,
.
Finally, differentiating u, we get
.
Substituting u, dv, v, and du into
using the form of v with the square root in the denominator only where it
will help us later,
Moving the constants out of the last integral, and expressing the integral of a sum as the sum of integrals, and then combining the exponents of x in all three integrals,
Now it is apparent that the original integral appears as part of the right-hand-side (did you see this coming?), so we will move it to the left side, combining it with the original integral, giving us
Now, if our main (or first) objective happened to be solving the indefinite integral using a recurrence relation, we might want to multiply both sides by 3/(2n-1). However, in this case, we are given a puzzle asking us to prove a particular recurrence relation involving the definite integral from 0 to 1. So we'll move the integral from the right-hand-side to the left-hand-side, and multiply through by 3:
Changing the indefinite to the definite,
Or, in other words,
Whew! A big sigh of relief. But we're not done! We need to tailor this factoid to give us an engine that will give us the answer sought by this problem.
Now, to start the recursion rolling, it's a matter of evaluating an integral we've already solved (the integral of dv, remember?) using the limits 0 and 1:
Now, applying the factoid we proved, for n=5,
And, finally, for n=8,
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