Problem 36 - AP Calculus with 8 Practice Tests 15th

Question

\(\int \frac{\tan ^{-1} y}{1+y^{2}} d y=\) (A) \(\left(\tan ^{-1} y\right)^{2}+C\) (B) \(\ln \left(1+y^{2}\right)+C\) (C) \(\ln \left(\tan ^{-1} y\right)+C\) (D) \(\frac{1}{2}\left(\tan ^{-1} y\right)^{2}+C\)

Step-by-Step Solution

Answer

The integral evaluates to \(\frac{1}{2}(\tan^{-1} y)^2 + C\), matching option (D).

Step 1: Analyze the Integral

We have the integral \( \int \frac{\tan^{-1} y}{1+y^{2}} \, dy \). Notice that the derivative of \( \tan^{-1} y \) is \( \frac{1}{1+y^2} \). This suggests that we can use substitution to solve the integral.

Step 2: Choose a Substitution

Let \( u = \tan^{-1} y \). Then, \( \frac{du}{dy} = \frac{1}{1+y^2} \). Thus, \( du = \frac{1}{1+y^2} \, dy \). This allows us to rewrite the integral in terms of \( u \).

Step 3: Substitute and Simplify

Substitute \( u \) and \( du \) into the integral. We get:\[ \int u \, du \]This is a simple integral of \( u \) with respect to \( u \).

Step 4: Integrate

Perform the integration:\[ \int u \, du = \frac{u^2}{2} + C \]Replace \( u \) with \( \tan^{-1} y \) to express the result in terms of the original variable.

Step 5: Replace Substitution and Identify Solution

Substitute back \( u = \tan^{-1} y \) into the integrated result:\[ \frac{u^2}{2} + C = \frac{1}{2} (\tan^{-1} y)^2 + C \]This expression matches option (D) of the given multiple choice.

Key Concepts

Inverse Trigonometric Functions

Inverse trigonometric functions are essential when dealing with integrals involving trigonometric expressions. These functions allow us to solve integrals which can appear complex at first sight. For example, in the given problem, we encounter the function \( \tan^{-1} y \), which is the inverse tangent. When dealing with inverse functions:

• The derivative of \( \tan^{-1} y \) is \( \frac{1}{1+y^2} \).
• This information is crucial as it helps identify substitutions or transformations to simplify the integral.

They often appear in calculus when resolving issues involving angles and their respective sides in right triangles. Additionally, these functions are bijective within their principal branch, providing precise relationships between angles and ratios. They offer an effective way to reverse the process used in calculating direct trigonometric values.

Definite and Indefinite Integrals

Integrals come in two forms: definite and indefinite. In our current problem, we deal with an indefinite integral. Indefinite integrals are used to find a general form of antiderivatives. They can be considered as the reverse process of taking a derivative. An indefinite integral:

• Includes an arbitrary constant of integration, represented by \( C \).
• Used when a specific function is needed without known limits.

When solving an indefinite integral, we seek to reverse the differentiation process to recover the original function that originated the derivative. In practical applications, understanding both definite and indefinite integrals is crucial, as they are extensively used in fields such as physics for calculating area, volume, and other fundamental properties. A good grasp of these concepts allows students to solve a wide range of mathematical problems effectively.

Substitution Method

The substitution method is a powerful tool for solving integrals, especially when the expression is complex or not in a standard form. This technique simplifies the integral by changing the variable of integration, making it easier to handle. In our exercise:

• We identify \( u = \tan^{-1} y \) as a suitable substitution.
• Then, relate \( du \, = \, \frac{1}{1+y^2} \, dy \) to transform the integral into a simpler form.

Through substitution, the given integral \( \int \frac{\tan^{-1} y}{1+y^{2}} \, dy \) becomes \( \int u \, du \), a straightforward form. This process not only simplifies calculations but also aids in recognizing patterns and relationships within integrals. The substitution method is highly effective when dealing with inverse trigonometric functions due to their specific derivative forms. By changing the perspective to a simpler variable, solving the integral becomes straightforward, leading to solutions that match the choices provided in multiple-choice problems like this one.