AP EAMCET · Maths · Indefinite Integration
\(\int e^{2 x+3} \sin 6 x d x=\)
- A \(\frac{e^{2 x+3}}{40}(2 \sin 6 x+6 \cos 6 x)+c\)
- B \(\frac{e^{2 x+3}}{40}(2 \cos 6 x+6 \sin 6 x)+c\)
- C \(\frac{e^{2 x+3}}{40}(\sin 6 x-3 \cos 6 x)+c\)
- D \(\frac{e^{2 x+3}}{40}(\cos 6 x-3 \sin 6 x)+c\)
Answer & Solution
Correct Answer
(C) \(\frac{e^{2 x+3}}{40}(\sin 6 x-3 \cos 6 x)+c\)
Step-by-step Solution
Detailed explanation
Let \(\mathrm{I}=\int e^{2 x+3} \sin 6 x d x\) \(=\sin 6 x \frac{e^{2 x+3}}{2}-\int\left(6 \cos 6 x \cdot \frac{e^{2 x+3}}{2}\right) d x\) \(=\frac{e^{2 x+3}}{2} \sin 6 x-3 \int e^{2 x+3} \cos 6 x d x\)…
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