To solve the integral:

$I = \int \frac{1}{(16 - t^2)^{3/2}} \, dt$

We can use a standard trigonometric substitution, which is effective for integrals of the form $\frac{1}{(a^2 - x^2)^{3/2}}$.

Step 1: Apply a trigonometric substitution.

Let $t = 4 \sin \theta$. This gives:

• $dt = 4 \cos \theta \, d\theta$
• $16 - t^2 = 16 - 16\sin^2 \theta = 16 \cos^2 \theta$

Thus, the integral becomes:

$I = \int \frac{1}{(16 \cos^2 \theta)^{3/2}} \cdot 4 \cos \theta \, d\theta$

Simplifying the powers:

$I = \int \frac{4 \cos \theta}{16^{3/2} \cos^3 \theta} \, d\theta = \int \frac{4}{16^{3/2} \cos^2 \theta} \, d\theta$

Since $16^{3/2} = 64$, the integral becomes:

$I = \frac{1}{64} \int \sec^2 \theta \, d\theta$

Step 2: Integrate.

We know that the integral of $\sec^2 \theta$ is $\tan \theta$. So:

$I = \frac{1}{64} \tan \theta + C$

Step 3: Substitute back in terms of $t$.

Since $t = 4 \sin \theta$, we can find $\tan \theta$ as:

$\sin \theta = \frac{t}{4}, \quad \cos \theta = \sqrt{1 - \sin^2 \theta} = \sqrt{1 - \left( \frac{t}{4} \right)^2} = \frac{\sqrt{16 - t^2}}{4}$

Thus:

$\tan \theta = \frac{\sin \theta}{\cos \theta} = \frac{t}{\sqrt{16 - t^2}}$

Therefore, the solution to the integral is:

$I = \frac{1}{64} \cdot \frac{t}{\sqrt{16 - t^2}} + C$

Final Answer:

$\int \frac{1}{(16 - t^2)^{3/2}} \, dt = \frac{t}{64 \sqrt{16 - t^2}} + C$

Let me know if you'd like more details or have any further questions!

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Here are some related questions you might find interesting:

1. How does trigonometric substitution work for integrals of the form $\int \frac{1}{(a^2 - x^2)^{3/2}} \, dx$?
2. What other substitution methods could be applied to integrals involving square roots?
3. How do we handle integrals of rational functions like $\frac{1}{(x^2 - a^2)^{3/2}}$?
4. Can the result of this integral be generalized for other powers besides $\frac{3}{2}$?
5. What is the role of the identity $\sec^2 \theta = 1 + \tan^2 \theta$ in simplifying trigonometric integrals?

Tip: When dealing with integrals involving square roots of quadratic expressions, always consider a trigonometric substitution, as it often simplifies the integral significantly.