Problem 27 - Thomas Calculus

Question

Evaluate the integrals by using a substitution prior to integration by parts. \(\int_{0}^{\pi / 3} x \tan ^{2} x d x\)

Step-by-Step Solution

Answer

\( \frac{\sqrt{3}\pi - 3}{4} \)

Step 1: Identify a Suitable Substitution

To simplify the integral, we'll use the substitution \( u = \tan(x) \). This implies \( du = \sec^2(x) \, dx \). Therefore, \( dx = \frac{du}{\sec^2(x)} \). We need to express \( x \tan^2(x) \, dx \) in terms of \( u \). Note that \( \tan^2(x) = u^2 \), so our substitution becomes \( x \tan^2(x) = xu^2 \). However, \( x \) cannot be directly replaced in terms of \( u \), thus we leave \( x \) for integration by parts.

Step 2: Change the Limits of Integration

Convert the limits of integration from \( x \) to \( u \). When \( x = 0 \), \( u = \tan(0) = 0 \). When \( x = \frac{\pi}{3} \), \( u = \tan(\frac{\pi}{3}) = \sqrt{3} \). Thus, the new limits will go from \( u = 0 \) to \( u = \sqrt{3} \).

Step 3: Substitute and Set Up for Integration by Parts

The integral becomes \( \int_0^{\sqrt{3}} x \frac{u^2}{\sec^2(x)} \, du \). Since \( \sec^2(x) = 1 + \tan^2(x) = 1 + u^2 \), the integral is written as \( \int_0^{\sqrt{3}} x \frac{u^2}{1+u^2} \, du \). Compute \( x = \arctan(u) \), yielding \( \int_0^{\sqrt{3}} \arctan(u) \frac{u^2}{1+u^2} \, du \). Now, set it up for integration by parts where \( v = \arctan(u) \) and \( dw = \frac{u^2}{1+u^2} \, du \).

Step 4: Determine the Components for Integration by Parts

Identify \( v = \arctan(u) \) leading to \( dv = \frac{1}{1+u^2} \, du \). Let \( w = \int \frac{u^2}{1+u^2} \, du \). This is a simple integral: let \( u^2 = w \), then \( dw = 2u \, du \) or \( \, du = \frac{dw}{2u} \). Substitute into the integral \( \int \frac{w}{1+w} \, \frac{dw}{2\sqrt{w}} \). Simplifying yields \( \int \frac{u}{2} \, du = \frac{u^2}{4} + C \).

Step 5: Integrate by Parts

The integration by parts formula is \( \int u \, dv = uv - \int v \, du \). Substitute to get \( \int \arctan(u) \frac{u^2}{1+u^2} \, du = \arctan(u) \cdot \frac{u^2}{2} \bigg|_0^{\sqrt{3}} - \int \frac{u^2}{2} \cdot \frac{1}{1+u^2} \, du = \frac{1}{2} \left( \sqrt{3} \cdot \arctan(\sqrt{3}) - 0 \right) - \frac{1}{2} \int_0^{\sqrt{3}} u \, du \).

Step 6: Evaluate the Remaining Simple Integral

From the previous integration by parts, evaluate the remaining integral \( \int_0^{\sqrt{3}} u \, du = \left[ \frac{u^2}{2} \right]_0^{\sqrt{3}} \). Solving this gives \( \frac{(\sqrt{3})^2}{2} = \frac{3}{2} \).

Step 7: Compute the Final Result

The expression now becomes \( \frac{1}{2}( \sqrt{3} \cdot \frac{\pi}{3}) - \frac{1}{2} \cdot \frac{3}{2} \). Simplify it to get the final answer: \( \frac{\sqrt{3}\pi - 3}{4} \).

Key Concepts

Integration by Parts

Integration by parts is a method used in calculus to solve integrals involving products of functions. It is particularly handy when dealing with integrals that are difficult to address using basic integration techniques. The formula for integration by parts is given by:
\[ \int u \, dv = uv - \int v \, du \]
This key formula lets us transform the original integral into a simpler one that is easier to solve.
Let's break it down:

• : This is a function that, when differentiated, simplifies the problem.
• dv: This is the remaining part of the integral, which simplifies when integrated.
• uv: The product of the two functions after application of the formula.
• \(\int v \, du\): The new integral that results from the process, hopefully easier to solve than the original.

In our example, we applied integration by parts to an integral with an arctan function component, strategically picking \(v = \arctan(u)\) and \(dw = \frac{u^2}{1+u^2} \, du\). The choice of functions is critical to effectively applying this method and simplifying the problem.

Substitution Method

The substitution method is a powerful tool in integration that simplifies integrals by changing variables. It works when you can write the integral's variable part in terms of another variable, typically to turn a complex expression into something more manageable.
The idea is similar to using the chain rule for differentiation in reverse.
In a nutshell:

• Identify a substitution that simplifies the integral. Ask yourself which part contains the overall complexity.
• Rewrite the integral in terms of the new variable. Do not forget to change the differential \(dx\) as well, using the derivative of your substitution.
• Change the limits of integration if you are working with definite integrals. This ensures computations are accurate under the new variable.

In the original problem, the substitution \( u = \tan(x) \) was used, simplifying the integrable expression. We transitioned from \(x\) to \(u\), thus narrowing down the scope of the integration task. This move turns an awkward trigonometric expression into a polynomial, further helping subsequent integration.

Trigonometric Integrals

Trigonometric integrals arise when the integrand involves trigonometric functions like sine, cosine, or tangent. These integrals can often appear daunting, given the oscillatory nature of trigonometric functions, but various methods, including substitution, can help tackle them.
Several strategies are developed:

• Express everything in terms of sine and cosine. This often makes the integrals more approachable.
• If \(\tan^2(x)\) appears, remember its trigonometric identity: \(\tan^2(x) = \sec^2(x) - 1\).
• Employ substitution techniques that simplify or convert trigonometric parts into algebraic expressions.

In the exercise provided, reevaluating \(\tan^2(x)\) using its identity and substituting accordingly were both used to simplify and set the stage for further application of integration techniques, showcasing the power of these methods in managing complex trigonometric forms.