AP EAMCET · Maths · Indefinite Integration
\(\int \frac{e^{\sin x}(\sin 2 x-8 \cos x)}{2(\sin x-3)^2} d x=\)
- A \(e^{\sin x}(\sin x-3)+c\)
- B \(\frac{e^{\sin x}}{(\sin x-3)^2}+c\)
- C \(e^{\sin x}(\sin x-3)^2+c\)
- D \(\frac{e^{\sin x}}{\sin x-3}+c\)
Answer & Solution
Correct Answer
(D) \(\frac{e^{\sin x}}{\sin x-3}+c\)
Step-by-step Solution
Detailed explanation
\( \int \frac{e^{\sin x}(2 \sin x \cos x-8 \cos x)}{2(\sin x-3)^2} d x = \int e^{\sin x} \cos x \frac{\sin x-4}{(\sin x-3)^2} d x \) Let \( u = \sin x \). Then \( d u = \cos x d x \). The integral becomes \( \int e^u \frac{u-4}{(u-3)^2} d u \).…
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