indefinite-integration

Question: $\int_{{}}^{{}}{{{e}^{-2x}}\sin 3x\ dx=}$



1) $\frac{1}{13}{{e}^{-2x}}[\sin 3x+\cos 3x]+c$
2) $-\frac{1}{13}{{e}^{-2x}}[\sin 3x+\cos 3x]+c$
3) $\frac{1}{13}{{e}^{-2x}}[2\sin 3x+3\cos 3x]+c$
4) $-\frac{1}{13}{{e}^{-2x}}[2\sin 3x+3\cos 3x]+c$
Solution: Explanation: No Explanation
Integration by parts

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