Problem 25 - Essential Calculus

Question

Find the indefinite integral and check the result by differentiation. $$ \int\left(\tan ^{2} y+1\right) d y $$

Step-by-Step Solution

Answer

The indefinite integral of \(\tan^{2} y + 1\) with respect to \(y\) is \(\tan y + y + C\).

Step 1: Integral Calculation

Use the trigonometric identity that says: \(\tan^2 y = \sec^{2} y - 1\). Thus the integral problem becomes: \[ \int (\sec^2 y) dy \]. Now we know that the integral of \(\sec^{2} y\) with respect to \(y\) is \(\tan y\). Thus, the antiderivative (integral) of \(\tan^{2} y + 1\) with respect to \(y\) is \(\tan y + y + C\). The \(C\) represents the constant of integration.

Step 2: Checking the Result

After finding the integral, we want to make sure that we did it correctly. So, we will differentiate our result and it should give us back the original function, \(\tan^{2} y + 1\). We take the derivative of \(\tan y + y + C\) which gives \(\sec^{2} y + 1\). Again using the identity \(\tan^2 y = \sec^2 y - 1\), we rewrite our result as \(\tan^{2} y + 1\), which matches the original function.

Key Concepts

Understanding Antiderivatives

When talking about antiderivatives, we are referring to functions that, when differentiated, yield the given function back again. In simpler words, finding an antiderivative involves solving an integral. In our example, the indefinite integral \( \int (\tan^2 y + 1) \ dy \) was found to be \( \tan y + y + C \). The \( C \) is crucial since it represents the constant of integration, meaning there is a family of possible solutions here. Each represents a vertical shift of the function because integration does not specify a particular starting point on the y-axis. Why is the concept of an antiderivative important? Here are a few reasons:

• Antiderivatives help solve definite integrals when limits are added.
• They help us describe accumulations and areas under curves.
• They're central in calculus for solving differential equations.

The process we use for calculating the indefinite integral \( \int ( an^2 y + 1) \ dy \) shows just how directly useful antiderivatives can be in checking results with differentiation.

Breaking Down Trigonometric Identities

Trigonometric identities are handy tools in calculus, especially for simplifying integrals and derivatives related to trigonometric functions. In our example, the identity \( \tan^2 y = \sec^2 y - 1 \) played a key role in transforming the integral. By recognizing this identity, we rewrite the integral \( \int (\tan^2 y + 1) \ dy \) as \( \int \sec^2 y \ dy \).But why are these identities crucial in calculus?

• They simplify expressions, making integration and differentiation more straightforward.
• They allow for easy substitution, an essential technique for solving integrals.
• They reduce complex fractions into simpler components.

Learning these identities provides a strong foundation in solving trigonometric problems, and in this example, it made solving the integral much simpler by relying on an identity rather than direct integration.

Differentiation as a Verification Tool

Differentiation is a process in calculus that involves finding a derivative, which represents the rate of change of a function. In the context of verifying the correctness of an indefinite integral solution, differentiation works by taking the derivative of the antiderivative. If the steps were completed correctly, differentiating should result in the original function.In our case, after calculating the indefinite integral \( \tan y + y + C \), we check the result. Differentiating \( \tan y + y + C \) gives us \( \sec^2 y + 1 \), and using the identity \( \tan^2 y = \sec^2 y - 1 \), it simplifies back to \( \tan^2 y + 1 \), matching the function we started with.Using differentiation as a check is useful for several reasons:

• It confirms the accuracy of integration results.
• Helps identify mistakes if the integration does not match the original function.
• Increases understanding of both integration and differentiation.

Thus, differentiation is not just a tool to find rates of change, but also serves as a vital step in validating calculus processes.