Part 1: Choosing the Trigonometric Substitution

Given the integral:
$I = \int_0^{\frac{5\sqrt{3}}{2}} \frac{dx}{\sqrt{25 - x^2}}$ we recognize the expression $\sqrt{25 - x^2}$, which suggests a sine substitution:
$x = 5\sin\theta$ Thus, the correct answer is C. $x = 5\sin\theta$.

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Part 2: Rewriting the Integral Using Substitution

Using the substitution $x = 5\sin\theta$, we differentiate both sides:

$dx = 5\cos\theta \, d\theta$

Also, substituting into the square root term:

$\sqrt{25 - x^2} = \sqrt{25 - 25\sin^2\theta} = \sqrt{25(1 - \sin^2\theta)} = \sqrt{25\cos^2\theta} = 5\cos\theta.$

Rewriting the integral in terms of $\theta$:

$I = \int_0^{\frac{5\sqrt{3}}{2}} \frac{5\cos\theta \, d\theta}{5\cos\theta}$

Since $5\cos\theta$ cancels in the numerator and denominator:

$I = \int d\theta.$

Now, we need to find the new limits in terms of $\theta$:

• When $x = 0$: $0 = 5\sin\theta \Rightarrow \sin\theta = 0 \Rightarrow \theta = 0.$

• When $x = \frac{5\sqrt{3}}{2}$: $\frac{5\sqrt{3}}{2} = 5\sin\theta \Rightarrow \sin\theta = \frac{\sqrt{3}}{2}.$ Since $\sin\theta = \frac{\sqrt{3}}{2}$ corresponds to $\theta = \frac{\pi}{3}$,

the integral becomes:

$I = \int_0^{\frac{\pi}{3}} d\theta.$

Let me know if you would like further clarifications!

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Related Questions

1. How do you evaluate the integral $\int \frac{dx}{\sqrt{a^2 - x^2}}$?
2. Why do we use trigonometric substitution in integrals involving square roots?
3. How can you verify the correctness of a substitution?
4. What are the three main trigonometric substitutions and when are they used?
5. Can we solve this integral using a different method, like a geometric approach?

Tip: Always check the limits carefully when changing variables in definite integrals!