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Chapter 5: Problem 61

find the given integral. \(\int \frac{\sinh x}{1+\cosh x} d x\)

Short Answer

Expert verified
The short answer is: \(\int \frac{\sinh x}{1+\cosh x} d x = \ln |1 + \cosh x| + C\).

Step by step solution

01

Substitute

Let's substitute \(u = 1 + \cosh x\). Now, we find the derivative of \(u\) with respect to \(x\): \( \frac{d u}{d x} = \frac{d (1 + \cosh x)}{d x} = \sinh x\).
02

Rewrite the integral

Now let's rewrite the integral in terms of \(u\). Since \(\frac{d u}{d x} = \sinh x\), we can write \(d u = \sinh x d x\). Thus, the given integral becomes: \(\int \frac{\sinh x}{1+\cosh x} d x = \int \frac{1}{u} d u\).
03

Integrate with respect to u

The integral \(\int \frac{1}{u} d u\) is a standard integral, and it is equal to \(\ln |u| + C\), where \(C\) is the integration constant.
04

Substitute back

Now we need to substitute back \(u = 1 + \cosh x\) into our result: \(\ln |u| + C = \ln |1 + \cosh x| + C\).
05

Final answer

Therefore, the integral of the given function is: \(\int \frac{\sinh x}{1+\cosh x} d x = \ln |1 + \cosh x| + C\).

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