Problem 207 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

\(\int_{-3 \pi / 2}^{-\pi / 2}\left[(x+\pi)^{3}+\cos ^{2}(x+3 \pi)\right] d x\) is equal to \(\quad\) [2006] (A) \(\frac{\pi^{4}}{32}\) (B) \(\frac{\pi^{4}}{32}+\frac{\pi}{2}\) (C) \(\frac{\pi}{2}\) (D) \(\frac{\pi}{4}-1\)

Step-by-Step Solution

Answer

The integral evaluates to \( \frac{\pi}{2} \), which is answer (C).

Step 1: Simplify the Problem

To solve the integral \( \int_{-3 \pi / 2}^{-\pi / 2}\left[(x+\pi)^{3}+\cos ^{2}(x+3 \pi)\right] d x \), first recognize that \(x+\pi\) and \(x+3\pi\) are simple translations. Define a substitution for clarity: let \( u = x + \pi \), then \( du = dx \). This changes the limits of integration: when \( x = -3\pi/2 \), \( u = -\pi/2 \), and when \( x = -\pi/2 \), \( u = \pi/2 \). Rewrite the integral as \( \int_{-\pi/2}^{\pi/2} \left[ u^3 + \cos^2(u + 2\pi) \right] du \). Since \( \cos(u + 2\pi) = \cos(u) \), the integral simplifies to: \( \int_{-\pi/2}^{\pi/2} \left[ u^3 + \cos^2(u) \right] du \).

Step 2: Evaluate the Integral of \(u^3\)

The integral of \( u^3 \) is easier due to symmetry because \( u^3 \) is an odd function about \( u = 0 \). Therefore, the integral \( \int_{-\pi/2}^{\pi/2} u^3 \, du = 0 \) because the area under the curve from \( -\pi/2 \) to \( 0 \) cancels out with the area from \( 0 \) to \( \pi/2 \).

Step 3: Evaluate the Integral of \(\cos^2(u)\)

Use the identity \( \cos^2(u) = \frac{1 + \cos(2u)}{2} \) to rewrite the integral. Thus, \( \int_{-\pi/2}^{\pi/2} \cos^2(u) \; du = \int_{-\pi/2}^{\pi/2} \frac{1}{2} \, du + \int_{-\pi/2}^{\pi/2} \frac{1}{2}\cos(2u) \, du \). The first part is \( \frac{1}{2} \times (\pi / 2 - (-\pi / 2)) = \frac{1}{2} \times \pi = \frac{\pi}{2} \). The second part is zero because \( \int_{-\pi/2}^{\pi/2} \cos(2u) \, du = 0 \) since \( \cos(2u) \) is also an even function about \( u = 0 \).

Step 4: Add Results from Step 2 and Step 3

Combine the results from Step 2 and Step 3. The integral for \( u^3 \) was \( 0 \) and the integral for \( \cos^2(u) \) was \( \frac{\pi}{2} \). Therefore, the total integral is \( 0 + \frac{\pi}{2} = \frac{\pi}{2} \).

Step 5: Match the Result to Given Options

The answer obtained from Steps 1 to 4 shows that the integral evaluates to \( \frac{\pi}{2} \), which corresponds to option (C).

Key Concepts

Trigonometric Substitution

Trigonometric substitution is a powerful technique in calculus to simplify the integration of functions involving square roots, particularly those that are part of trigonometric identities. In our exercise, we applied a different substitution, changing variables from \(x\) to \(u = x + \pi\). This translation aligns with the behaviour of trigonometric functions across their periodic intervals. Yet, true trigonometric substitution often involves using identities like \(\sin^2\theta + \cos^2\theta = 1\) to replace variables in integrands. For example, we might use \(x = \sin \theta\) to simplify integrals that include \(\sqrt{1-x^2}\). This isn't exactly what was needed for the original problem, but understanding it can help in more complex integration scenarios involving radicals and trig functions. Here, recognizing that \(\cos(u + 2\pi) = \cos(u)\) used the basics of trigonometric periodicity to simplify what could have been a more challenging calculation.

Odd and Even Functions

Understanding odd and even functions is crucial when working with definite integrals over symmetric intervals, such as \([-a, a]\). An **odd function** satisfies \(f(-x) = -f(x)\). Its integral over a symmetric interval results in zero because the positive and negative areas cancel each other out. In this exercise, we dealt with \(u^3\), which is an odd function. This quickly simplified that part of the integral to zero.
On the other hand, **even functions** satisfy \(f(-x) = f(x)\), meaning their symmetric parts around the y-axis are identical, thus doubling the value of the distinct half. For instance, \(\cos^2(u)\) is even, meaning the integration over the interval contributes to a non-zero result. Recognizing these properties can save computation time and lead to simplified solutions. Moreover, knowing when a function is odd or even helps in making appropriate substitutions to reveal these properties more clearly.

Integration by Substitution

Integration by substitution is akin to the reverse process of the chain rule for derivatives. It aids in transforming an integral into a more recognizable form. In the provided exercise, a simple substitution \(u = x + \pi\) made it easier to handle the integration limits and simplified the cosine function due to its periodic properties. By switching to a new variable, we effectively shifted the problem to evaluate the integral from \(-\pi/2\) to \(\pi/2\). This technique is not just limited to variable shifts but can be used creatively with different functions. For instance, changing from \(t\) to \( \theta \) using trigonometric substitutions, or even using exponential functions for hyperbolic identities. The goal is to make the integral easier or possible to calculate, so understanding substitution is vital for solving complex integrals effectively.