A) \[{{e}^{2x}}\sin x+c\]
B) \[-{{e}^{2x}}\sin x+c\]
C) \[-{{e}^{2x}}\cos x+c\]
D) \[{{e}^{2x}}\cos x+c\]
Correct Answer: D
Solution :
\[\int_{{}}^{{}}{{{e}^{2x}}(-\sin x+2\cos x)\,dx}\] \[=-\int_{{}}^{{}}{{{e}^{2x}}\sin x\,dx}+2\int_{{}}^{{}}{{{e}^{2x}}\cos x\,dx}\] \[={{e}^{2x}}\cos x-2\int_{{}}^{{}}{{{e}^{2x}}\cos x\,dx+2\int_{{}}^{{}}{{{e}^{2x}}\cos x\,dx+c}}\] \[={{e}^{2x}}\cos x+c.\] Aliter : \[\int_{{}}^{{}}{{{e}^{2x}}(2\cos x-\sin x)\,dx}={{e}^{2x}}\cos x+c\] \[\left\{ \because \,\,\,\int_{{}}^{{}}{{{e}^{kx}}\left\{ k\,f(x)+{f}'(x) \right\}dx={{e}^{kx}}f(x)+c} \right\}\]
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