Evaluating the Integral of ln|sin(u)| + c Using Integration by Parts

ln|sinu|+c

To evaluate the integral of ln|sin(u)| + c, we can use integration by parts

To evaluate the integral of ln|sin(u)| + c, we can use integration by parts.

Let’s start by writing the integral in a form that can be more easily integrated. The absolute value in ln|sin(u)| can be written as a piecewise function:

ln|sin(u)| = ln(sin(u)), if sin(u) > 0
ln(-sin(u)), if sin(u) < 0 The natural logarithm function ln(x) can only be applied to positive values, so we need to consider the different cases separately. Case 1: sin(u) > 0

In this case, we can simply integrate ln(sin(u)) as follows:
∫ ln(sin(u)) du

Now, let’s apply integration by parts:
Let:
u = ln(sin(u))
dv = du
du = (1/sin(u)) * cos(u) du
v = ∫ dv = u

Using the integration by parts formula:
∫ u dv = uv – ∫ v du

We can now substitute the values into the formula:
∫ ln(sin(u)) du = u * ln(sin(u)) – ∫ u * (1/sin(u)) * cos(u) du

Simplifying the integral on the right-hand side:
∫ u * (1/sin(u)) * cos(u) du = ∫ u * cot(u) du

Case 2: sin(u) < 0 In this case, we need to evaluate ∫ ln(-sin(u)) du differently. We can make a substitution to convert the integral into a positive value. Let v = -u, then u = -v and du = -dv. Now we have: ∫ ln(-sin(u)) du = -∫ ln(sin(v)) (-dv) = ∫ ln(sin(v)) dv So, in this case, we have the same integral as in Case 1: ∫ ln(sin(u)) du. Now, combining the results from both cases, we have: ∫ ln|sin(u)| du = u * ln(sin(u)) - ∫ u * cot(u) du + C Where C represents the constant of integration.

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