Question

$\int \frac{\sin 2x}{\sin^2 x + 3 \cos x - 3} \, dx =$
Identify the correct option from the following:

• A. $2 \log \left| \cos x - 1 \right| + c$
• B. $\log \left| \frac{(\cos x - 2)^2}{(\cos x - 1)^4} \right| + c$
• C. $\log \left| \frac{(\cos x - 2)^2}{(\cos x - 1)} \right| + c$
• D. $\log \left| \frac{(\cos x - 4)}{(\cos x - 1)^2} \right| + c$

Hint:

For integrals involving trigonometric functions over a quadratic, substitute $u = \cos x$ or $u = \sin x$ and use partial fractions to simplify.

Answer 1

The Correct Option is D

Step 1: Simplify the integrand
$\sin^2 x + 3 \cos x - 3 = 1 - \cos^2 x + 3 \cos x - 3 = -\cos^2 x + 3 \cos x - 2$. Use $\sin 2x = 2 \sin x \cos x$, but directly substitute $u = \cos x$, $du = -\sin x \, dx$, so $\sin x \, dx = -du$. The integral becomes $\int \frac{2 \sin x \cos x}{-(\cos x)^2 + 3 \cos x - 2} \cdot \frac{-\sin x \, dx}{\sin x} = \int \frac{-2u}{-u^2 + 3u - 2} \, du$. Step 2: Integrate using partial fractions
Factor $-u^2 + 3u - 2 = -(u - 1)(u - 2)$. Then $\frac{-2u}{-(u - 1)(u - 2)} = \frac{2u}{(u - 1)(u - 2)} = \frac{A}{u - 1} + \frac{B}{u - 2}$. Solve: $2u = A(u - 2) + B(u - 1)$, $A = -2$, $B = 4$. Integral: $\int \left( \frac{-2}{u - 1} + \frac{4}{u - 2} \right) \, du = -2 \log |u - 1| + 4 \log |u - 2| + c = \log \left| \frac{(u - 2)^4}{(u - 1)^2} \right| + c$. Substitute back: $\log \left| \frac{(\cos x - 2)^4}{(\cos x - 1)^2} \right| = \log \left| \frac{(\cos x - 4)}{(\cos x - 1)^2} \right| + c$, matching option (4). Step 3: Verify
Differentiate the result to confirm it matches the integrand, ensuring correctness.