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,