The main idea is to exploit the trigonometric identity,
sin²(θ) + cos²(θ) = 1
2. For an integral containing 16 - 81x², you might substitute x = 4/9 sin(θ) (with differential dx = 4/9 cos(θ) dθ, but without an actual integral to work with this isn't really important). Then
16 - 81x² = 16 - 81 (4/9 sin(θ))²
… = 16 - 81 (16/81 sin²(θ))
… = 16 - 16 sin²(θ)
… = 16 (1 - sin²(θ))
… = 16 cos²(θ)
so that in the root expression, we would end up with
[tex]\left(16 - 81x^2\right)^{7/2} = \left(16\cos^2(\theta)\right)^{7/2} = 2^{14} |\cos(\theta)|^7[/tex]
since [tex](ab)^c=a^cb^c[/tex] for all real a, b, and c; [tex]16^{7/2}=\left(2^4\right)^{7/2}=2^{14}[/tex]; and [tex]\sqrt{x^2}=|x|[/tex] for all real x.
The goal is to replace x with some multiple of sin(θ) that makes the coefficients factor out like they did here, which then lets you reduce 1 - sin²(θ) to cos²(θ).
And don't be discouraged by the absolute values; in the context of a definite integral, there are things that can be done to remove them or otherwise simplify absolute value expressions.
3. Substitute z = 1/√8 sin(θ) (so that dz = 1/√8 cos(θ) dθ). Then
1 - 8z² = 1 - 8 (1/√8 sin(θ))²
… = 1 - 8 (1/8 sin²(θ))
… = 1 - sin²(θ)
… = cos²(θ)
so that
[tex]\left(1-8z^2\right)^{3/2} = \left(\cos^2(\theta)\right)^{3/2} = |\cos(\theta)|^3[/tex]
