Find \(\int \frac{x}{\sqrt{4-x^{2}}} \mathrm{~d} x\).

Short Answer

Expert verified
The final solution of the integral is \(\int \frac{x}{\sqrt{4-x^{2}}} \mathrm{~d} x = -\sqrt{4 - x^2} + C\), where C is the constant of integration.

Step by step solution

01

Choose an appropriate trigonometric substitution

To make the denominator simpler, we can use the identity \(1 - \sin^2(\theta) = \cos^2(\theta)\). We can choose a trigonometric substitution such that \(4 - x^2 = 4\cos^2(\theta)\). Let \(x = 2\sin(\theta)\), then \(\mathrm{d} x = 2\cos(\theta) \mathrm{~d}\theta\).
02

Substitute and simplify the integrand

Now substitute \(x = 2\sin(\theta)\) and \(\mathrm{d} x = 2\cos(\theta) \mathrm{~d}\theta\) in the integral and simplify the integrand: \begin{align*} \int \frac{x}{\sqrt{4-x^{2}}} \mathrm{~d} x &= \int \frac{2\sin(\theta)}{\sqrt{4 - (2\sin(\theta))^2}} (2\cos(\theta) \mathrm{~d}\theta) \\ &= \int \frac{4\sin(\theta)\cos(\theta)}{\sqrt{4(1 - \sin^2(\theta))}} \mathrm{~d}\theta \\ &= \int \frac{4\sin(\theta)\cos(\theta)}{2\sqrt{\cos^2(\theta)}} \mathrm{~d}\theta \\ &= \int 2\sin(\theta) \mathrm{~d}\theta \end{align*}
03

Evaluate the integral in terms of \(\theta\)

The integral has now become simpler. Evaluate the integral with respect to \(\theta\). $$ \int 2\sin(\theta) \mathrm{~d}\theta = -2\cos(\theta) + C $$ where C is the constant of integration.
04

Convert back to \(x\)

Now we substitute \(\theta\) back in terms of \(x\). We know \(x = 2\sin(\theta)\). Then, \(\sin(\theta) = \frac{x}{2}\). Since \(\sin^2(\theta) + \cos^2(\theta) = 1\), we can find \(\cos(\theta)\): $$ \cos^2(\theta) = 1 - \sin^2(\theta) = 1 - \left(\frac{x}{2}\right)^2 = \frac{4 - x^2}{4} $$ So, \(\cos(\theta) = \frac{\sqrt{4 - x^2}}{2}\). Finally, substitute \(\cos(\theta)\) back into the expression for the integral: $$ -2\cos(\theta) + C = -2\left(\frac{\sqrt{4 - x^2}}{2}\right) + C = -\sqrt{4 - x^2} + C $$ Thus, the final solution is: $$ \int \frac{x}{\sqrt{4-x^{2}}} \mathrm{~d} x = -\sqrt{4 - x^2} + C $$

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