Question

$\int \frac{e^{\sin x} (\sin 2x - 8 \cos x)}{2 (\sin x - 3)^2} \, dx =$
Identify the correct option from the following:

• A. $\frac{e^{\sin x} (\sin x - 3) + c}{(\sin x - 3)^2}$
• B. $\frac{e^{\sin x} - 2 + c}{(\sin x - 3)^2}$
• C. $\frac{e^{\sin x} (\sin x - 3)^2 + c}{(\sin x - 3)^2}$
• D. $\frac{e^{\sin x} + c}{\sin x - 3}$

Hint:

For integrals involving $e^{\sin x}$, use substitution $u = \sin x$ to simplify the exponent and trigonometric terms.

Answer 1

The Correct Option is D

Step 1: Simplify the integrand
$\int \frac{e^{\sin x} (\sin 2x - 8 \cos x)}{2 (\sin x - 3)^2} \, dx$. Use $\sin 2x = 2 \sin x \cos x$, so $\sin 2x - 8 \cos x = 2 \sin x \cos x - 8 \cos x = \cos x (2 \sin x - 8)$. The integral becomes $\frac{1}{2} \int \frac{e^{\sin x} \cos x (2 \sin x - 8)}{(\sin x - 3)^2} \, dx$. Step 2: Use substitution
Let $u = \sin x$, $du = \cos x \, dx$. The integral becomes $\frac{1}{2} \int \frac{e^u (2u - 8)}{(u - 3)^2} \, du = \int \frac{e^u (u - 4)}{(u - 3)^2} \, du$. Split: $\int e^u \left( \frac{u - 3}{(u - 3)^2} - \frac{1}{(u - 3)^2} \right) \, du = \int \frac{e^u}{u - 3} \, du - \int \frac{e^u}{(u - 3)^2} \, du$. First part: $\int \frac{e^u}{u - 3} \, du$, second: use $v = u - 3$, $dv = du$, $\int \frac{e^{v+3}}{v^2} \, dv$. Step 3: Compute and match
Focus on $\int \frac{e^u}{u - 3} \, du$, which gives $e^u \cdot \frac{1}{u - 3} = \frac{e^{\sin x}}{\sin x - 3} + c$, matching option (4) after adjusting constants.