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

Question

\(\int_{-3 \pi / 2}^{-\pi / 2}\left[(x+\pi)^{3}+\cos ^{2}(x+3 \pi)\right] d x\) is equal to \(\quad\) (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 is \( \frac{\pi}{2} \), so the answer is (C).

Step 1: Simplifying the Cosine Term

First, simplify the cosine term inside the integral. Note that \( \cos^2(x + 3\pi) = \cos^2(x) \) because \( \cos(x + 3\pi) = -\cos(x) \) and squaring removes the sign change.

Step 2: Setting up the Integral

Substitute the simplified expression into the integral: \[ \int_{-3\pi/2}^{-\pi/2} \left((x+\pi)^3 + \cos^2(x)\right) dx \].

Step 3: Splitting the Integral

Separate the integral into two parts: \[ \int_{-3\pi/2}^{-\pi/2} (x+\pi)^3 dx + \int_{-3\pi/2}^{-\pi/2} \cos^2(x) dx \].

Step 4: Evaluating \((x+\pi)^3\) Integral

Evaluate \( \int (x+\pi)^3 dx \) using the power rule. The antiderivative is \( \frac{1}{4}(x+\pi)^4 \). Evaluate from \(-3\pi/2\) to \(-\pi/2\): \[ \frac{1}{4}\left((-\pi/2+\pi)^4 - (-3\pi/2+\pi)^4\right) \].

Step 5: Calculating Limits for \((x+\pi)^3\)

Calculate the evaluated limits: \((-\pi/2+\pi)^4 = (\pi/2)^4 = \pi^4 / 16\) and \((-3\pi/2+\pi)^4 = (-\pi/2)^4 = \pi^4 / 16\). Their difference is zero.

Step 6: Evaluating \(\cos^2(x)\) Integral

Use the identity \( \cos^2(x) = \frac{1 + \cos(2x)}{2} \) to write \[ \int \cos^2(x) dx = \int \frac{1}{2} dx + \int \frac{1}{2} \cos(2x) dx \].

Step 7: Solving \(\cos^2(x)\) Part

Evaluate \( \int_{-3\pi/2}^{-\pi/2} \frac{1}{2} dx = \frac{1}{2} \cdot [-\pi/2 - (-3\pi/2)] = \frac{1}{2} \cdot \pi = \frac{\pi}{2} \). The term with \( \cos(2x) \) is zero due to symmetric limits.

Step 8: Combining Results

Add the results from the two parts of the integral: for \((x+\pi)^3\) it's zero; for \(\cos^2(x)\) it's \(\frac{\pi}{2}\). So, the final integral evaluates to \(\frac{\pi}{2}\).

Key Concepts

Cosine Squared Function

The cosine squared function, denoted as \( \cos^2(x) \), often appears in various integrals due to its periodic nature and symmetry. In trigonometry, an important identity is utilized to simplify integration involving \( \cos^2(x) \):

• \( \cos^2(x) = \frac{1 + \cos(2x)}{2} \)

This identity allows us to break the cosine squared function into two simpler terms. The first term, \( \frac{1}{2} \), is a constant, simplifying the integration process. The second term, \( \frac{1}{2}\cos(2x) \), can also be integrated more easily since it is a cosine function. When integrating over a specific interval, particularly symmetric limits like from \(-a\) to \(a\), some components of periodic functions like \( \cos(2x) \) can cancel out.
This simplification is what helps make solving integrals involving \( \cos^2(x) \) more straightforward and approachable. Thus, knowing these fundamental transformations and properties significantly eases the process of evaluating definite integrals.

Power Rule for Integration

The power rule for integration is one of the foundational techniques in calculus that allows the integration of polynomial terms. This rule is especially handy for terms of the form \( x^n \). Here’s the rule in a nutshell:

• If \( n eq -1 \), then \( \int x^n \, dx = \frac{x^{n+1}}{n+1} + C \), where \( C \) is the constant of integration.

In the context of definite integrals, the power rule is used before applying the limits of integration. For example, consider the expression \( (x+\pi)^3 \). First, we apply the power rule to find its antiderivative:

• \( \int (x+\pi)^3 \, dx = \frac{1}{4}(x+\pi)^4 + C \)

Then, to evaluate the definite integral, we plug in the upper and lower limits of the interval. In our problem, calculating these limits and their difference allows obtaining the integral's value. The power rule simplifies any polynomial integration, making it possible to tackle polynomial expressions easily within integrals.

Symmetric Limits in Integrals

Integrals involving symmetric limits, like from \(-a\) to \(a\), offer unique simplifications under certain conditions, particularly with periodic functions. When dealing with symmetric limits, some parts of the integral might cancel out based on the function’s properties.

• If a function is odd, such as \( f(-x) = -f(x) \), the integral from \(-a\) to \(a\) is zero.
• If a function is even, like the cosine squared function, this symmetry can simplify the calculations, although integrals over symmetric limits for even functions need extra consideration of periodic properties.

This feature of definite integrals harnesses the inherent symmetry of functions. For instance, this property is utilized with \( \cos(2x) \) because this function has cyclical behavior, which often results in cancellation over a symmetrically set boundary.
When evaluating complex integrals, recognizing these symmetric properties can often lead to simplification and save significant computation effort, as it did in the original exercise where the integration of the cosine term over symmetric limits simplified to a zero contribution.