Problem 61 - Single Variable Calculus

Question

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

Step-by-Step Solution

Answer

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

Step 1: Substitute

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

Step 2: 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: \(\int \frac{\sinh x}{1+\cosh x} d x = \int \frac{1}{u} d u\).

Step 3: 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.

Step 4: Substitute back

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

Step 5: Final answer

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