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Chapter 7: Problem 7

Evaluate the following integrals. $$\int_{0}^{5 / 2} \frac{d x}{\sqrt{25-x^{2}}}$$

Short Answer

Expert verified
Question: Evaluate the definite integral \(\int_{0}^{5/2}\frac{dx}{\sqrt{25-x^2}}\). Answer: The value of the definite integral is \(\frac{\pi}{6}\).

Step by step solution

01

1. Identify Trigonometric Substitution

We can utilize the trigonometric substitution \(x = 5\sin(\theta)\) because \(25 - x^2 = 25(1 - \sin^2(\theta)) = 25\cos^2(\theta)\). So, the substitution is: $$x = 5\sin(\theta)\Rightarrow dx = 5\cos(\theta)d\theta$$
02

2. Substitute Trig Functions and DX

Replace \(x\) and \(dx\) in the integral with the trigonometric substitution we found in the previous step, and simplify the integrand: $$\int_{0}^{5/2}\frac{dx}{\sqrt{25-x^2}} = \int_{\theta_1}^{\theta_2}\frac{5\cos(\theta)d\theta}{\sqrt{25(1-\sin^2(\theta))}}$$ Now, we need to find the new limits of integration, i.e., \(\theta_1\) and \(\theta_2\). At lower limit, when \(x=0\), $$0 = 5\sin(\theta_1) \Rightarrow \theta_1 = 0$$ At upper limit, when \(x=5/2\), $$\frac{5}{2} = 5\sin(\theta_2) \Rightarrow \theta_2 = \arcsin(\frac{1}{2}) = \frac{\pi}{6}$$ So the integral becomes: $$\int_0^{\pi/6}\frac{5\cos(\theta)d\theta}{\sqrt{25\cos^2(\theta)}}$$
03

3. Simplify the Integral

Now, let's simplify the integrand: $$\int_0^{\pi/6}\frac{5\cos(\theta)d\theta}{5\cos(\theta)}$$ By canceling common factors, we get: $$\int_0^{\pi/6} d\theta$$
04

4. Find the Antiderivative

The antiderivative of \(1\) with respect to \(\theta\) is simply: $$\int_0^{\pi/6} d\theta = \theta$$
05

5. Evaluate the Limits of Integration

Now, we will evaluate the antiderivative with the new limits of integration: $$\theta \bigg|_0^{\pi/6} = \left(\frac{\pi}{6}\right) - (0) = \frac{\pi}{6}$$ Thus, the value of the integral is: $$\int_{0}^{5 / 2} \frac{d x}{\sqrt{25-x^{2}}} = \frac{\pi}{6}$$

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