To evaluate the indefinite integral ∫(x²)(36-64x²)³/2 dx, we can use the trigonometric substitution method. The three basic trigonometric substitutions commonly used are:
1. Substitution 1: x = a sin θ
2. Substitution 2: x = a tan θ
3. Substitution 3: x = a sec θ
Let's choose Substitution 1: x = a sin θ.
To find the value of a, we can set 36-64x² equal to a perfect square. In this case, 36-64x² = 16(1-4x²). To make it a perfect square, we can set it equal to (4a sin θ)². Solving for a, we find that a = 2.
Now, we substitute x = 2 sin θ into the integral:
∫(x²)(36-64x²)³/2 dx
= ∫((2 sin θ)²)(36-64(2 sin θ)²)³/2 (2 cos θ) dθ
= 16∫sin² θ(1-4sin² θ)³/2 cos θ dθ
Using the trigonometric identity sin² θ = 1 - cos² θ, we can rewrite the integral as:
16∫(1-cos² θ)(1-4(1-cos² θ))³/2 cos θ dθ
= 16∫(1-cos² θ)(1-4+4cos² θ)³/2 cos θ dθ
= 16∫(1-3cos² θ)(1+4cos² θ)³/2 cos θ dθ
Now, we can expand the expression (1+4cos² θ)³/2:
16∫(1-3cos² θ)(1+4cos² θ)³/2 cos θ dθ
= 16∫(1-3cos² θ)(1+6cos² θ + 12cos⁴ θ + 8cos⁶ θ) cos θ dθ
Expanding further and distributing, we get:
16∫(cos θ - 3cos³ θ + 6cos⁵ θ - 3cos⁷ θ + 12cos² θ - 36cos⁴ θ + 24cos⁶ θ - 8cos⁸ θ) dθ
Now, we can integrate each term separately. Remember to use the trigonometric identities for integration. After integrating, you'll get the final result in terms of x.
It's important to note that the above steps may require further simplification and algebraic manipulation.