Integral Setup

We are asked to find the value of coefficient \(A\) given the integral equation:

\( \int \frac{\sin\theta d\theta}{(2+\cos\theta)(3+4\cos\theta)} = A \ln|2+\cos\theta| + B \ln|3+4\cos\theta| \)

Substitution Method

To solve the integral, we use the substitution:

Let \(u = \cos\theta\). Then, differentiating both sides gives \(du = -\sin\theta d\theta\).

Substituting these into the integral, we get:

\( \int \frac{-du}{(2+u)(3+4u)} \)

Partial Fraction Decomposition

We decompose the rational function \(\frac{-1}{(2+u)(3+4u)}\) using partial fractions:

\( \frac{-1}{(2+u)(3+4u)} = \frac{P}{2+u} + \frac{Q}{3+4u} \)

Multiplying both sides by \((2+u)(3+4u)\) gives:

\( -1 = P(3+4u) + Q(2+u) \)

To find \(P\), set \(u = -2\):

\( -1 = P(3+4(-2)) + Q(2-2) \)

\( -1 = P(3-8) \implies -1 = -5P \implies P = \frac{1}{5} \)

To find \(Q\), set \(u = -3/4\):

\( -1 = P(3+4(-3/4)) + Q(2-3/4) \)

\( -1 = P(0) + Q(\frac{8}{4}-\frac{3}{4}) \implies -1 = Q(\frac{5}{4}) \implies Q = -\frac{4}{5} \)

So the decomposition is:

\( \frac{-1}{(2+u)(3+4u)} = \frac{1/5}{2+u} + \frac{-4/5}{3+4u} \)

Integration Process

Now, we integrate the decomposed form:

\( \int \left( \frac{1/5}{2+u} + \frac{-4/5}{3+4u} \right) du \)

\( = \frac{1}{5} \int \frac{1}{2+u} du - \frac{4}{5} \int \frac{1}{3+4u} du \)

Integrating each term:

\( \int \frac{1}{2+u} du = \ln|2+u| \)

\( \int \frac{1}{3+4u} du = \frac{1}{4} \ln|3+4u| \)

Combining these results:

\( \frac{1}{5} (\ln|2+u|) - \frac{4}{5} \left( \frac{1}{4} \ln|3+4u| \right) + C \)

\( = \frac{1}{5} \ln|2+u| - \frac{1}{5} \ln|3+4u| + C \)

Determining Coefficient A

Substitute back \(u = \cos\theta\):

\( \frac{1}{5} \ln|2+\cos\theta| - \frac{1}{5} \ln|3+4\cos\theta| + C \)

Compare this result with the given form \(A \ln|2+\cos\theta| + B \ln|3+4\cos\theta|\).

By matching the coefficients of \(\ln|2+\cos\theta|\), we find:

\( A = \frac{1}{5} \)