Problem 35 - Calculus Early Transcendentals

Question

Determine each indefinite integral. \(\int \tanh ^{2} x d x\) (Hint: Use an identity.)

Step-by-Step Solution

Answer

Answer: The indefinite integral of \(\tanh^2 x\) with respect to \(x\) is \(x -\frac{1}{\cosh x} + C\), where \(C\) is the constant of integration.

Step 1: Rewrite using the identity

We begin by rewriting the given function using the identity mentioned above. The integral becomes: $$\int \tanh^2 x \; dx = \int \frac{\sinh^2 x}{\cosh^2 x} \; dx$$

Step 2: Use the properties of \(\cosh^2 x - \sinh^2 x = 1\)

Another useful identity is \(\cosh^2 x - \sinh^2 x = 1\). Using this identity, we can rewrite the function in the integral as follows: $$\int \frac{\sinh^2 x}{\cosh^2 x} \; dx = \int \frac{\cosh^2 x - 1}{\cosh^2 x} \; dx = \int \left(1 - \frac{1}{\cosh^2 x}\right) dx$$

Step 3: Break integral into two parts

Now, we can break the integral into two separate parts: $$\int \left(1 - \frac{1}{\cosh^2 x}\right) dx = \int 1 \; dx - \int \frac{1}{\cosh^2 x} \; dx$$

Step 4: Integrate the first part

The first integral can be easily integrated: $$\int 1 \; dx = x$$

Step 5: Integrate the second part

To integrate the second part, we need to use a substitution. Let \(u = \cosh x\), so \(du = \sinh x \; dx\). Thus, we get: $$\int \frac{1}{\cosh^2 x} \; dx = \int \frac{du}{u^2} = -\frac{1}{u}$$ Now, we back-substitute \(u\): $$-\frac{1}{u} = -\frac{1}{\cosh x}$$

Step 6: Combine the two parts and write the final answer

Now, we combine the two parts to write the final answer for the given integral: $$\int \tanh^2 x \; dx = \int \left(1 - \frac{1}{\cosh^2 x}\right)dx = x -\frac{1}{\cosh x} + C$$ Where \(C\) is the constant of integration.