Question

Find: \[ \int \frac{\cos x}{(2+\sin x)(4+\sin x)}\,dx. \]

Hint:

When linear terms in the denominator differ by a constant, you can skip full partial fractions by using the difference observation: \( \frac{1}{(t+2)(t+4)} = \frac{1}{2} \left[ \frac{1}{t+2} - \frac{1}{t+4} \right] \).

Answer 1

Concept: This problem can be solved effectively by the method of substitution followed by partial fraction decomposition. Since the derivative of \( \sin x \) is \( \cos x \), choosing \( t = \sin x \) converts the trigonometric integrand into a rational algebraic function.

Step 1: Apply integration by substitution.
Let: \[ t = \sin x \] Differentiating both sides with respect to \( x \): \[ dt = \cos x \, dx \] Substituting these values back into the given integral: \[ I = \int \frac{\cos x \, dx}{(2 + \sin x)(4 + \sin x)} = \int \frac{dt}{(2 + t)(4 + t)} \]

Step 2: Decomposition using partial fractions.
We can split the integrand using partial fractions: \[ \frac{1}{(2 + t)(4 + t)} = \frac{A}{2 + t} + \frac{B}{4 + t} \] Multiplying through by the common denominator \( (2 + t)(4 + t) \): \[ 1 = A(4 + t) + B(2 + t) \] To find the coefficients \( A \) and \( B \), we substitute the roots of the linear factors:
• Let \( t = -2 \): \( 1 = A(4 - 2) + B(0) \Rightarrow 1 = 2A \Rightarrow A = \frac{1}{2} \)
• Let \( t = -4 \): \( 1 = A(0) + B(2 - 4) \Rightarrow 1 = -2B \Rightarrow B = -\frac{1}{2} \) Thus, our integrand can be rewritten as: \[ \frac{1}{(2 + t)(4 + t)} = \frac{1}{2(2 + t)} - \frac{1}{2(4 + t)} \]

Step 3: Perform individual integrations.
Substitute the partial fractions back into the integral: \[ I = \int \left[ \frac{1}{2(2 + t)} - \frac{1}{2(4 + t)} \right] dt = \frac{1}{2} \log|2 + t| - \frac{1}{2} \log|4 + t| + C \] Using logarithm properties \( \log M - \log N = \log\left(\frac{M}{N}\right) \): \[ I = \frac{1}{2} \log\left| \frac{2 + t}{4 + t} \right| + C \]

Step 4: Substitute back original variables.
Replacing \( t \) with \( \sin x \) gives the final result: \[ I = \frac{1}{2} \log\left| \frac{2 + \sin x}{4 + \sin x} \right| + C \]