5. Let
. Note that we want this variable change to be reversible, so we tacitly assume 0 ≤ θ ≤ π/2. Then

and
. So the integral transforms to

Reduce the power by writing

Now let
, so that
. Then

Replace the variable to get the antiderivative back in terms of x and we have




6. Let
and
. It follows that

since, like in the previous integral, under this reversible variable change we assume -π/2 < θ < π/2. Over this interval, sec(θ) is positive.
Now,

The denominator reduces to

and so

Rewrite sin³(θ) just like before,

and substitute
again to get

Put everything back in terms of x :



2(b). For some constants a, b, c, and d, we have

3(a). For some constants a, b, and c,

5(a). For some constants a-f,

where we use the sum-of-5th-powers identity,
