Question

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

Short answer

Question: Evaluate the integral $$\int \frac{dx}{\sqrt{(x-1)(3-x)}}$$. Answer: The integral evaluates to $$\arcsin(x-2)+C$$.

Step-by-step solution

Completing the Square

First, let's complete the square for the quadratic function in the denominator: $$(x-1)(3-x) = -x^2+4x-3$$ Now complete the square by adding and subtracting the square of half of the coefficient of x: $$-x^2+4x-3=-\left(x^2-4x\right)-3=-\left(x^2-4x+\left(\frac{4}{2}\right)^2-\left(\frac{4}{2}\right)^2\right)-3=-\left(\left(x-2\right)^2-4\right)-3=-\left(x-2\right)^2+1$$ So, we can rewrite the integral as: $$\int \frac{dx}{\sqrt{1- (x-2)^2}}$$

Substitution

Now, let's make a substitution to solve the integral. We can see that the integrand resembles the derivative of arcsin. Let's make the following substitution: $$\sin \theta=x-2$$ With this substitution, the integral becomes: $$\int \frac{d\theta}{\sqrt{1-\sin^2\theta}}$$ Also, we have \(dx = \cos\theta d\theta\). Therefore, the integral becomes: $$\int \frac{\cos\theta d\theta}{\sqrt{1-\sin^2\theta}}$$ The expression inside the square root is equal to \(\cos^2\theta\), so we get: $$\int \frac{\cos\theta d\theta}{\cos\theta}$$ Now, everything inside the integral cancels out, and we are left with: $$\int d\theta$$

Evaluating the Integral

Now let's evaluate the remaining integral: $$\int d\theta=\theta+C$$ Where \(C\) is the constant of integration.

Back Substitution

Finally, let's substitute back to get our final answer. Since we made the substitution \(\sin \theta=x-2\), we have \(\theta=\arcsin(x-2)\). Therefore, the integral evaluates to: $$\int \frac{dx}{\sqrt{(x-1)(3-x)}}=\arcsin(x-2)+C$$

Key concepts

Integration Techniques

Integration techniques are essential tools in calculus for solving a variety of integrals. These techniques help simplify integrals into forms that are easier to evaluate. Understanding different integration methods allows you to effectively tackle complex problems, especially when dealing with integrals that don't fit standard formulas.

* **Substitution Method**: This technique involves changing variables to transform the integral into a simpler form. It’s beneficial for functions that are compositions, such as \(f(g(x))\). You might often see it used alongside completing the square, as it allows easy integration after simplifying the integrand.
* **Integration by Parts**: This is based on the product rule for differentiation and is useful when you have products of functions, like \(u(x)v(x)\). It requires identifying one function to differentiate and another to integrate within the integral.
* **Partial Fraction Decomposition**: Useful when integrating rational expressions, where the integrand is a ratio of two polynomials. It involves breaking down complex fractions into simpler parts that are easier to integrate separately.

Each technique has its own process and is suitable for different types of integrals. Choosing the right technique depends on the form and complexity of the integrand.

Trigonometric Substitution

Trigonometric substitution is a powerful method for evaluating integrals that involve square roots of quadratic expressions. This technique involves substituting a trigonometric function for the variable in an integral, making use of trigonometric identities to simplify the integrand.

In our solved problem, substitution was crucial, thanks to the form \(\sqrt{1 - (x-2)^2}\), which is reminiscent of the identity \(1 - \sin^2 \theta = \cos^2 \theta\). Rewriting the square root in terms of \(\cos^2 \theta\) allows us to simplify the integral dramatically.

Here's how you typically approach trigonometric substitution:
* For expressions like \(\sqrt{a^2 - x^2}\), substitute \(x = a \sin \theta\) and use \(dx = a \cos \theta d\theta\).
* For \(\sqrt{x^2 + a^2}\), use \(x = a \tan \theta\) with \(dx = a \sec^2 \theta d\theta\).
* For \(\sqrt{x^2 - a^2}\), choose \(x = a \sec \theta\) with \(dx = a \sec\theta \tan\theta d\theta\).

By utilizing these substitutions, the integral becomes easier to evaluate through standard trigonometric functions, paving the way to a rapid solution with fewer computational steps.

Definite and Indefinite Integrals

Understanding definite and indefinite integrals is crucial in calculus. These concepts relate to different aspects of integration and have unique properties and applications.

**Indefinite Integrals**: These integrals represent a family of functions and include an arbitrary constant \(C\). They are fundamental to solving differential equations and are written in the form \(\int f(x) \, dx = F(x) + C\). In our exercise, the solution \(\arcsin(x-2) + C\) is an example of an indefinite integral, meaning it represents a family of antiderivatives.

**Definite Integrals**: In contrast, these integrals calculate the net area under a curve between two points \(a\) and \(b\) on the x-axis. They do not include the constant \(C\). Instead, they are evaluated as \(\int_{a}^{b} f(x) \, dx = F(b) - F(a)\) using the Fundamental Theorem of Calculus. This theorem connects differentiation with integration, providing a reliable method to evaluate definite integrals by finding antiderivatives.

Recognizing the difference allows you to apply the correct approach in various scenarios, whether you're solving for a general solution or computing a specific numerical value.