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Chapter 6: Problem 35

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

Short Answer

Expert verified
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 by step solution

01

Rewrite using the identity

We begin by rewriting the given function using the identity mentioned above. The integral becomes: $$\int \tanh^2 x \; dx = \int \frac{\sinh^2 x}{\cosh^2 x} \; dx$$
02

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: $$\int \frac{\sinh^2 x}{\cosh^2 x} \; dx = \int \frac{\cosh^2 x - 1}{\cosh^2 x} \; dx = \int \left(1 - \frac{1}{\cosh^2 x}\right) dx$$
03

Break integral into two parts

Now, we can break the integral into two separate parts: $$\int \left(1 - \frac{1}{\cosh^2 x}\right) dx = \int 1 \; dx - \int \frac{1}{\cosh^2 x} \; dx$$
04

Integrate the first part

The first integral can be easily integrated: $$\int 1 \; dx = x$$
05

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 \(u\): $$-\frac{1}{u} = -\frac{1}{\cosh x}$$
06

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.

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