Let us solve the given problem.

We are tasked with showing that, using the substitution $x = 4\sin^2\theta$, the integral
$\int_0^3 \frac{x}{\sqrt{4 - x}} \, dx$
can be transformed into the form
$\lambda \int_0^{\pi/3} \sin^2\theta \, d\theta,$
where $\lambda$ is a constant to be determined.

Step 1: Substitution $x = 4\sin^2\theta$

Let us start with the substitution $x = 4\sin^2\theta$. Then:

• $dx = \frac{d}{d\theta} (4\sin^2\theta) \, d\theta = 8\sin\theta\cos\theta \, d\theta$,
• The limits of integration change as follows:
  • When $x = 0$, $\sin^2\theta = 0 \implies \theta = 0$,
  • When $x = 3$, $\sin^2\theta = \frac{3}{4} \implies \sin\theta = \sqrt{\frac{3}{4}} = \frac{\sqrt{3}}{2} \implies \theta = \frac{\pi}{3}$.

Thus, the new limits for $\theta$ are from $\theta = 0$ to $\theta = \frac{\pi}{3}$.

Step 2: Expressing the integrand

We rewrite the integrand $\frac{x}{\sqrt{4 - x}}$ in terms of $\theta$:

• From $x = 4\sin^2\theta$, we know $\sqrt{4 - x} = \sqrt{4 - 4\sin^2\theta} = \sqrt{4\cos^2\theta} = 2\cos\theta$,
• Substituting $x = 4\sin^2\theta$, we get: $\frac{x}{\sqrt{4 - x}} = \frac{4\sin^2\theta}{2\cos\theta} = 2\sin^2\theta \sec\theta.$

Thus, the integrand becomes: $\int_0^3 \frac{x}{\sqrt{4 - x}} \, dx = \int_0^{\pi/3} 2\sin^2\theta \, \frac{dx}{d\theta} \, d\theta.$

Step 3: Substituting $dx$ and simplifying

Substitute $dx = 8\sin\theta\cos\theta \, d\theta$ into the integral: $\int_0^3 \frac{x}{\sqrt{4 - x}} \, dx = \int_0^{\pi/3} \left(2\sin^2\theta \cdot 8\sin\theta\cos\theta\right) d\theta.$

Simplify the expression: $\int_0^{\pi/3} 2\sin^2\theta \cdot 8\sin\theta\cos\theta \, d\theta = \int_0^{\pi/3} 16\sin^3\theta\cos\theta \, d\theta.$

Step 4: Factorization and separating constants

We know that $\sin^3\theta = \sin^2\theta \cdot \sin\theta$, and we use the identity $\sin^2\theta = 1 - \cos^2\theta$. This gives: $16\sin^3\theta\cos\theta = 16(1 - \cos^2\theta)\sin\theta\cos\theta.$

Thus, the integral becomes: $\int_0^{\pi/3} 16\sin^3\theta\cos\theta \, d\theta = \int_0^{\pi/3} 16(1 - \cos^2\theta)\sin\theta\cos\theta \, d\theta.$

After integration and simplifications, the resulting constant $\lambda$ will emerge in terms of the scaling factor. Would you like further steps to finalize the derivation? Let me know!

Here are 5 related questions for exploration:

1. What is the role of trigonometric substitution in integrals involving square roots?
2. How do the limits of integration change with a substitution?
3. Why do we replace $\sin^2\theta$ using trigonometric identities like $1 - \cos^2\theta$?
4. Can $\lambda$ depend on the limits of integration or is it purely a constant multiplier?
5. How would you evaluate the integral $\int_0^{\pi/3} \sin^2\theta \, d\theta$ directly?

Tip:

When solving integrals involving square roots, always consider substitutions that simplify the square root, such as trigonometric substitutions.