Problem 3 - Essential Calculus

Question

Use differentiation to match the antiderivative with the correct integral. [Integrals are labeled (a), (b), (c), and \((\mathbf{d}) .]\) (a) \(\int \sin x \tan ^{2} x d x\) (b) \(8 \int \cos ^{4} x d x\) (c) \(\int \sin x \sec ^{2} x d x\) (d) \(\int \tan ^{4} x d x\) $$ y=x-\tan x+\frac{1}{3} \tan ^{3} x $$

Step-by-Step Solution

Answer

The derivative matches with Integral (c). Hence, the antiderivative \(y = x - \tan x + \frac{1}{3}\tan^3 x\) matches with Integral (c) \(\int \sin x \sec^2 x dx\).

Step 1: Differentiate the antiderivative

Differentiate \(y = x - \tan x + \frac{1}{3}\tan^3 x\). \n\nTo do this, we apply the power and chain rule of differentiation: \n\n\(\frac{dy}{dx} = 1 - \sec^2 x + \tan^2 x \sec^2 x\)

Step 2: Simplify the derivative

We simplify the derivative to be able to relate it back to the original integrals. To simplify, we observe that \(\tan^2 x \sec^2 x\) is the same as \(\sec^2 x \tan^2 x\). Hence, we write the derivative as: \n\n\(\frac{dy}{dx} = 1 - \sec^2 x + \sec^2 x \tan^2 x\)

Step 3: Match the derivative with the correct integral

It's time to match the derivative with the integral. Using the fundamental theorem of calculus, if \(\frac{dy}{dx} = f(x)\), then \(\int f(x) dx = y\). Hence: \n\nIntegral (a) \(\int \sin x \tan^2 x dx\), Integral (b) \(8 \int \cos^4 x dx\), Integral (c) \(\int \sin x \sec^2 x dx\) or Integral (d) \(\int \tan^4 x dx\) must be equal to \(y = x - \tan x + \frac{1}{3}\tan^3 x\).\n\nAfter comparing the derivative with the integrals, we find that the derivative matches Integral (c) \(\int \sin x \sec^2 x dx\). Integral (c) and the derivative both contain \(\sec^2 x\) and \(\tan^2 x\) terms, and hence are equivalent after integral evaluation according to the fundamental theorem of calculus.

Key Concepts

Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus stands as a bridge between differentiation and integration, two core operations in calculus. It connects the process of differentiation with that of integration and, in essence, provides a way to evaluate definite integrals easily.

This theorem is split into two parts: the first part essentially tells us that if we have a continuous function defined over an interval, an antiderivative of this function can help us calculate the definite integral over that interval. The second part asserts that differentiation is the reverse process of integration. That is, if you have a function that was obtained through integration, differentiating it will yield the original function used in the integration process.

Applying this concept to the exercise, once we differentiate the given antiderivative, we should end up with the function that, when integrated, gives us the same antiderivative. By following the Fundamental Theorem of Calculus, we can match the derivative with the correct integral by comparing the resulting function after differentiation with the integrals provided in the options.

Power Rule of Differentiation

Differentiation can often be streamlined with the use of rules and the power rule is one of the simplest and most powerful among them. The power rule states that for any function of the form \(f(x) = ax^n\) where \(a\) and \(n\) are constants, the derivative of \(f\) with respect to \(x\) is \(f'(x) = anx^{(n-1)}\).

For example, in the given exercise, to differentiate \(\frac{1}{3}tan^3 x\), we apply the power rule and get \(tan^2 x sec^2 x\) as a part of the derivative. This rule simplifies the process of differentiation, especially when dealing with polynomial functions or functions that can be rewritten in the power form.

However, when functions involve trigonometric identities, like in the exercise, we must employ the power rule with care and sometimes combine it with other rules, like the chain rule, to get the correct derivative.

Chain Rule of Differentiation

When functions are composed of other functions, the chain rule is the strategy we use to find their derivatives. The chain rule states that if \(h(x) = f(g(x))\), then the derivative of \(h\) with respect to \(x\) is \(h'(x) = f'(g(x)) \cdot g'(x)\). In simpler terms, it says to differentiate the outer function, multiply by the derivative of the inner function.

In the context of the exercise, when we differentiate \(\tan x\) in the antiderivative, we consider \(\tan x\) as \(\sin x / \cos x\) or \(\sin x \cdot \sec x\) – a product of two functions. So, we apply the chain rule combined with the product rule. We first treat \(\tan x\) as the outer function and \(x\) as the inner function, obtaining \(\sec^2 x\) as the derivative of the outer function and \(1\) as the derivative of the inner function. Multiplying these gives us the part of the derivative \( - \sec^2 x\) from the antiderivative. Using the chain rule correctly is crucial for handling complex functions and ensuring accurate differentiation.