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Chapter 7: Problem 3

What change of variables is suggested by an integral containing \(\sqrt{100-x^{2}} ?\)

Short Answer

Expert verified
Question: Solve the following integral: $$\int \sqrt{100 - x^2} dx$$ Answer: $$\int \sqrt{100 - x^2} dx = 50\arcsin{\frac{x}{10}} + 25\sin{2\arcsin{\frac{x}{10}}} + C$$

Step by step solution

01

Setting up the substitution

To simplify the integral, we will use the substitution \(x = 10 \sin{\theta}\). This substitution helps us to rewrite the expression under the square root in terms of \(\theta\). Next, we need to find the derivative of this substitution with respect to \(\theta\): $$\frac{dx}{d\theta} = 10 \cos{\theta}$$
02

Rewrite the square root term

Now, we will replace \(x\) with the substitution we've set up earlier: \(x = 10 \sin{\theta}\). This enables us to rewrite the square root expression as follows: $$\sqrt{100 - x^2} = \sqrt{100 - (10\sin{\theta})^2} = \sqrt{100 - 100 \sin^2{\theta}} = \sqrt{100(1 - \sin^2{\theta})}$$ Recall the Pythagorean identity: \(1 - \sin^2{\theta} = \cos^2{\theta}\). We can use this identity to further simplify the expression under the square root: $$\sqrt{100(1 - \sin^2{\theta})} = \sqrt{100\cos^2{\theta}} = 10\cos{\theta}$$
03

Rewrite the integral

With the simplifications performed in the previous steps, we can now rewrite the original integral in terms of \(\theta\): $$\int \sqrt{100 - x^2} dx = \int 10\cos{\theta} \cdot 10\cos{\theta} d\theta = 100\int \cos^2{\theta} d\theta$$
04

Solving the integral

To solve the integral \(100\int \cos^2{\theta} d\theta\), we can use the identity \(\cos^2{\theta} = \frac{1+\cos{2\theta}}{2}\) to simplify the integrand before integrating: $$100\int \cos^2{\theta} d\theta = 100\int \frac{1+\cos{2\theta}}{2} d\theta = 50\int (1+\cos{2\theta}) d\theta$$ Now, we can solve the integral: $$50\int (1+\cos{2\theta}) d\theta = 50(\int 1 d\theta + \int \cos{2\theta} d\theta) = 50(\theta + \frac{\sin{2\theta}}{2}) + C = 50\theta + 25\sin{2\theta} + C$$
05

Convert back to x

Finally, we need to rewrite the result in terms of the original variable \(x\). We know that \(x = 10\sin{\theta}\), so we can solve for \(\theta\): $$\theta = \arcsin{\frac{x}{10}}$$ Now, substitute this back into the result of the integral: $$50\theta + 25\sin{2\theta} + C = 50\arcsin{\frac{x}{10}} + 25\sin{2\arcsin{\frac{x}{10}}} + C$$ Thus, the integral, considering the substitution suggested by the expression \(\sqrt{100-x^2}\) is: $$\int \sqrt{100 - x^2} dx = 50\arcsin{\frac{x}{10}} + 25\sin{2\arcsin{\frac{x}{10}}} + C$$

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