Question

Find: \[ \int \frac{(3 \cos x - 2) \sin x}{5 - \sin^2 x - 4 \cos x} \, dx. \]

Hint:

When solving trigonometric integrals, use appropriate substitutions such as \( u = \cos x \) or \( u = \sin x \) to simplify the given expression.

Answer 1

Step 1: Simplify the denominator. The denominator of the given integral is: \[ 5 - \sin^2 x - 4 \cos x. \] Using the identity \( \sin^2 x = 1 - \cos^2 x \), rewrite the expression: \[ 5 - (1 - \cos^2 x) - 4 \cos x = \cos^2 x - 4 \cos x + 4. \] Thus, the integral simplifies to: \[ \int \frac{(3 \cos x - 2) \sin x}{\cos^2 x - 4 \cos x + 4} \, dx. \] Step 2: Use substitution. Let \( u = \cos x \), then: \[ du = -\sin x \, dx. \] Rewriting the integral in terms of \( u \): \[ -\int \frac{3u - 2}{u^2 - 4u + 4} \, du. \] Since the denominator can be factored as: \[ u^2 - 4u + 4 = (u - 2)^2, \] the integral becomes: \[ -\int \frac{3u - 2}{(u - 2)^2} \, du. \] Step 3: Simplify the fraction. Rewrite the numerator: \[ 3u - 2 = 3(u - 2) + 4. \] Thus, the integral splits into: \[ -\int \frac{3(u - 2)}{(u - 2)^2} \, du - \int \frac{4}{(u - 2)^2} \, du. \] Step 4: Evaluate the integrals. For the first term: \[ -\int \frac{3(u - 2)}{(u - 2)^2} \, du = -\int \frac{3}{u - 2} \, du = -3 \ln |u - 2|. \] For the second term: \[ -\int \frac{4}{(u - 2)^2} \, du = \frac{4}{u - 2}. \] Step 5: Substitute back \( u = \cos x \). \[ -3 \ln |\cos x - 2| + \frac{4}{\cos x - 2} + C. \] Final Answer: \[ \int \frac{(3 \cos x - 2) \sin x}{5 - \sin^2 x - 4 \cos x} \, dx = -3 \ln |\cos x - 2| + \frac{4}{\cos x - 2} + C. \]