Question

Evaluate the following integrals. Consider completing the square. $$\int \frac{d x}{\sqrt{(x-1)(3-x)}}$$

Short answer

Question: Evaluate the integral $$\int \frac{d x}{\sqrt{(x-1)(3-x)}}$$ Answer: $$\arcsin(x-2) + C$$

Step-by-step solution

Complete the square

Let's rewrite the expression inside the square root as a completed square. This can be helpful for simplifying the expression and making it easier to integrate. To complete the square, look for a square that has the same zeroes as the original expression. $$\sqrt{(x-1)(3-x)} = \sqrt{-(x^2-4x+3)} = \sqrt{-(x^2-4x+4-4+3)} = \sqrt{-(x-2)^2+1}$$ Now, the integral becomes: $$\int \frac{d x}{\sqrt{-(x-2)^2+1}}$$

Substitution

Now, let's perform a substitution that will simplify the integral. By observation, we can notice that the expression in the square root resembles the Pythagorean identity \(\sin^2 \theta + \cos^2 \theta = 1\). Let's use the substitution \(y = x - 2\), so \(x = y + 2\) and \(d x = d y\). Then, the expression becomes: $$\int \frac{d y}{\sqrt{-y^2+1}}$$ Now, we can perform another substitution, \(y = \sin \theta\). The derivative of \(y\) with respect to \(\theta\) is \(d y = \cos \theta d \theta\). Thus, the integral becomes: $$\int \frac{\cos \theta d \theta}{\sqrt{1-\sin^2 \theta}}$$ Now, the expression "-sin^2(θ)" in the denominator becomes: $$\sqrt{1-\sin^2 \theta} = \sqrt{\cos^2 \theta} = \cos \theta$$ Therefore, the integral becomes: $$\int \frac{\cos \theta d \theta}{\cos \theta}$$

Evaluate the Integral

Now, the integral is simple to solve, as the cosine terms cancel out: $$\int \frac{\cos \theta d \theta}{\cos \theta} = \int d \theta$$ The integral of \(d \theta\) is simply \(\theta + C\), where C is the integration constant. Now we need to substitute back to our original variable, \(x\). From before, we have \(y = x-2\) which implies \(x=y+2\). And we have \(y = \sin \theta\), so we have \(x-2 = \sin \theta\), which means: $$\theta = \arcsin(x-2)$$ Thus, the final result is: $$\arcsin(x-2) + C$$